The initial pressure in the vessel.

Question

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Answer 1

Question

Chapter 12, Problem 72AP

(a)

To determine

The initial pressure in the vessel.

(a)

Expert Solution

Answer to Problem 72AP

the initial pressure in the vessel is

$1.00×103 Pa$ .

Explanation of Solution

Section1:

To determine: The constant volume occupied by the gas.

Answer: The volume of the gas is $0.524 m3$ .

Explanation:

Given Info:

The radius of the spherical vessel is $0.500 m$ .

Formula to calculate the volume of the gas is,

$V=43πr3$

r is the radius of the vessel

Substitute $0.500 m$ for r to find the volume.

$VSun=43π(0.500 m)3=0.524 m3$

Thus, volume of the gas is $0.524 m3$ .

Conclusion:

From section 1,

The volume of the gas is $0.524 m3$

The amount of hydrogen in the container is $14.4 mol$ .

The amount of oxygen in the container is $7.2 mol$ .

The initial temperature is $20 °C$

To find the initial pressure of the vessel, consider the components in the gas mixture in container as ideal gases. Hence, the total pressure the mixture exerts on the container will be the sum of pressure of the components of the mixture

The pressure exerted by the hydrogen molecule is,

$PH2=nH2RTV$ (I)

$nH2$ is the amount of hydrogen molecules in moles
R is the universal gas constant
T is the absolute temperature
V is the volume of the container

The pressure exerted by the oxygen molecule is,

$PO2=nO2RTV$ (II)

$nO2$ is the amount of oxygen molecule in moles

The total pressure exerted on the container is the sum of equation (I) and (II).

Therefore,

$P=PH2+PO2=(nH2+nO2)RTV$

Substitute $14.4 mol$ for $nH2$ , $7.2 mol$ for $nO2$ , $8.31 J/mol⋅K$ for R, $20 °C$ for T and $0.524 m3$ for V to find the initial pressure of the container,

$P=(14.4 mol+7.2 mol)(8.31 J/mol⋅K)((20+273)K)0.524 m3=1.00×103 Pa$

Thus, the initial pressure in the vessel is $1.00×103 Pa$ .

(b)

To determine

The initial internal energy of the gas.

(b)

Expert Solution

Answer to Problem 72AP

The initial internal energy of the gas is

$1.31×105 J$

Explanation of Solution

Given Info:

The amount of hydrogen in the container is $14.4 mol$ .

The amount of oxygen in the container is $7.2 mol$ .

The molar specific heat of hydrogen at constant volume is $20.4 J/mol.K$ .

The molar specific heat of oxygen at constant volume is $21.1 J/mol⋅K$ .

The internal energy of the gas is the sum of the internal energy of the components of the gas mixture.

Formula to calculate the internal energy of the gas is,

$U=UH2,i+UO2,i=(nO2Cv,o2+nH2Cv,H2)Ti$

$nO2$ is the amount of oxygen in moles
$Cv,o2$ is the molar specific heat of oxygen in constant volume
$nH2$ is the amount of hydrogen in moles
$Cv,H2$ is the molar specific heat of hydrogen in constant volume

Substitute $14.4 mol$ for $nH2$ , $20.4 J/mol.K$ for $Cv,H2$ , $7.2 mol$ $nO2$ and $21.1 J/mol⋅K$ for $Cv,o2$ to find the initial internal energy of the gas,

$Ui=|(14.4 mol)(20.4 J/mol⋅K)+(7.2 mol)(21.1 J/mol⋅S)|=1.31×105 J$

Thus, the initial internal energy of the gas is $1.31×105 J$

Conclusion:

The initial internal energy of the gas is $1.31×105 J$

(c)

To determine

The energy produced if the gas burns out to water vapor completely.

(c)

Expert Solution

Answer to Problem 72AP

The energy produced if the gas burns out to water vapor completely is

$3.48×106 J$ .

Explanation of Solution

Given Info:

The amount of hydrogen in the container is $14.4 mol$ .

The amount of oxygen in the container is $7.2 mol$ .

The initial temperature is $20 °C$

If the gas burns out to water vapor completely, instead of each one mole of hydrogen; one mole water will produce. In the container since the volume is constant, no other work done will be there. The water produced is equal to the amount of hydrogen.

Thus, the amount of water produced is $14.4 mol$ .

The chemical energy required to produce one mole of water is $241.8 kJ$ .

Formula to calculate the energy produced in combustion is,

$ΔU=(241.8 kJ/mol)n$

n is the number of moles of water produced

Substitute $14.4 mol$ for n to find the energy produced,

$ΔU=(241.8 kJ/mol)(14.4 mol)=3.48×106 J$

Thus, the energy produced is $3.48×106 J$ .

Conclusion:

The energy produced if the gas burns out to water vapor completely is $3.48×106 J$ .

(d)

To determine

The temperature and pressure of the steam.

(d)

Expert Solution

Answer to Problem 72AP

The temperature and pressure of the water vapor is

$9.28×103 K$ and

$2.12×106 Pa$ .

Explanation of Solution

Section1:

To determine: The temperature of the steam

Answer: The final temperature of the steam is $9.28×103 K$

Explanation:

Given Info:

The molar heat capacity for water vapor is $27.0 J/mol⋅K$ .

The initial internal energy of the gas is $1.31×105 J$ .

The energy produced if the gas burns out to water vapor completely is $3.48×106 J$

For an ideal gas,

Formula to calculate the temperature of the steam is,

$Tf=UfnwCv,w$ (I)

$Uf$ is the final internal energy
$nw$ is the amount of water in moles
$Cv,w$ is the molar heat capacity of water

Since, after the combustion; the finial internal energy is,

$Uf=Ui+ΔU$

Thus, the equation (I) is given by,

$Tf=Ui+ΔUnwCv,w$

Substitute $1.31×105 J$ for $Ui$ , $3.48×106 J$ for $ΔU$ , $14.4 mol$ for $nw$ and $27.0 J/mol⋅K$ for $Cv,w$ to find the temperature of the steam,

$T=(1.31×105 J)(3.48×106 J)(14.4 mol)(27.0 J/mol⋅K)=9.28×103 K$

Thus, the final temperature of the steam is $9.28×103 K$

Section2:

To determine: The pressure of the steam

Answer: The pressure of the steam is $2.12×106 Pa$

Explanation:

Given Info:

From section1,

The final temperature of the steam is $9.28×103 K$

For an ideal gas,

Formula to calculate the pressure of the steam is,

$P=nwRTV$

$nw$ is the amount of water vapor in moles
R is the universal gas constant
T is the final temperature
V is the volume occupied by the gas

Substitute $14.4 mol$ for $nw$ , $8.31 J/mol⋅K$ for R, $9.28×103 K$ for T and $0.524 m3$ for V to find the pressure of the steam,

$P=(14.4 mol)(8.31 J/mol⋅K)(9.28×103 K)0.524 m3=2.12×106 Pa$

Thus, the pressure of the steam is $2.12×106 Pa$

Conclusion:

The temperature and pressure of the water vapor is $9.28×103 K$ and $2.12×106 Pa$ .

(e)

To determine

The mass of steam and steam’s density.

(e)

Expert Solution

Answer to Problem 72AP

the mass of steam and steams density is

$0.259 kg$ and

$0.494 kg/m3$ .

Explanation of Solution

Section1:

To determine: The mass of steam.

Answer: The mass of the steam is $0.259 kg$ .

Explanation:

Given Info:

The mass of water is $18.0×10−3 kg/mol$

The amount of steam is $14.4 mol$

Formula to calculate the mass of the steam after the combustion is,

$ms=nsMw$

$ns$ is the amount of steam in moles
$Mw$ is the mass of water in moles

Substitute $14.4 mol$ for $ns$ and $18.0×10−3 kg/mol$ for $Mw$ to find the mass of steam,

$ms=(14.4 mol)(18.0×10−3 kg/mol)=0.259 kg$

Thus, mass of the steam is $0.259 kg$ .

Conclusion:

From section1,

Mass of the steam is $0.259 kg$ .

The volume of the gas is $0.524 m3$ .

Formula to calculate the density of the steam is,

$ρs=msV$

$ms$ is the mass of the steam
V is the volume of the gas

Substitute $0.259 kg$ for $ms$ , $0.524 m3$ for V to find the density of the steam,

$ρs=0.259 kg0.524 m3=0.494 kg/m3$

Thus, the steam’s density is $0.494 kg/m3$ .

(f)

To determine

The initial exhaust velocity of the exhaust steam.

(f)

Expert Solution

Answer to Problem 72AP

The initial exhaust velocity of the exhaust steam is

$2.93×103 m/s$ .

Explanation of Solution

Given Info:

The Bernoulli’s equation is given by,

$P2+12ρ2v22+ρsgh2=P1+12ρsv12+ρsgh1$

Assume that the steam in the container is essentially at rest.

Thus,

$v1=0$ , $P2=0$ and $h2=h1$ .

Formula to calculate the exhaust velocity is,

$v2=2P1ρs$

$P1$ is the initial pressure of the gas
$ρs$ is the density of the steam

Substitute $2.12×106 Pa$ for $P1$ and $0.494 kg/m3$ for $ρs$ to find the exhaust velocity,

$v2=2(2.12×106 Pa)0.494 kg/m3=2.93×103 m/s$

Thus, the exhaust velocity is $2.93×103 m/s$ .

Conclusion:

The initial exhaust velocity of the exhaust steam is $2.93×103 m/s$ .

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Answer 2

Question

Chapter 12, Problem 72AP

(a)

To determine

The initial pressure in the vessel.

(a)

Expert Solution

Answer to Problem 72AP

the initial pressure in the vessel is

$1.00×103 Pa$ .

Explanation of Solution

Section1:

To determine: The constant volume occupied by the gas.

Answer: The volume of the gas is $0.524 m3$ .

Explanation:

Given Info:

The radius of the spherical vessel is $0.500 m$ .

Formula to calculate the volume of the gas is,

$V=43πr3$

r is the radius of the vessel

Substitute $0.500 m$ for r to find the volume.

$VSun=43π(0.500 m)3=0.524 m3$

Thus, volume of the gas is $0.524 m3$ .

Conclusion:

From section 1,

The volume of the gas is $0.524 m3$

The amount of hydrogen in the container is $14.4 mol$ .

The amount of oxygen in the container is $7.2 mol$ .

The initial temperature is $20 °C$

To find the initial pressure of the vessel, consider the components in the gas mixture in container as ideal gases. Hence, the total pressure the mixture exerts on the container will be the sum of pressure of the components of the mixture

The pressure exerted by the hydrogen molecule is,

$PH2=nH2RTV$ (I)

$nH2$ is the amount of hydrogen molecules in moles
R is the universal gas constant
T is the absolute temperature
V is the volume of the container

The pressure exerted by the oxygen molecule is,

$PO2=nO2RTV$ (II)

$nO2$ is the amount of oxygen molecule in moles

The total pressure exerted on the container is the sum of equation (I) and (II).

Therefore,

$P=PH2+PO2=(nH2+nO2)RTV$

Substitute $14.4 mol$ for $nH2$ , $7.2 mol$ for $nO2$ , $8.31 J/mol⋅K$ for R, $20 °C$ for T and $0.524 m3$ for V to find the initial pressure of the container,

$P=(14.4 mol+7.2 mol)(8.31 J/mol⋅K)((20+273)K)0.524 m3=1.00×103 Pa$

Thus, the initial pressure in the vessel is $1.00×103 Pa$ .

(b)

To determine

The initial internal energy of the gas.

(b)

Expert Solution

Answer to Problem 72AP

The initial internal energy of the gas is

$1.31×105 J$

Explanation of Solution

Given Info:

The amount of hydrogen in the container is $14.4 mol$ .

The amount of oxygen in the container is $7.2 mol$ .

The molar specific heat of hydrogen at constant volume is $20.4 J/mol.K$ .

The molar specific heat of oxygen at constant volume is $21.1 J/mol⋅K$ .

The internal energy of the gas is the sum of the internal energy of the components of the gas mixture.

Formula to calculate the internal energy of the gas is,

$U=UH2,i+UO2,i=(nO2Cv,o2+nH2Cv,H2)Ti$

$nO2$ is the amount of oxygen in moles
$Cv,o2$ is the molar specific heat of oxygen in constant volume
$nH2$ is the amount of hydrogen in moles
$Cv,H2$ is the molar specific heat of hydrogen in constant volume

Substitute $14.4 mol$ for $nH2$ , $20.4 J/mol.K$ for $Cv,H2$ , $7.2 mol$ $nO2$ and $21.1 J/mol⋅K$ for $Cv,o2$ to find the initial internal energy of the gas,

$Ui=|(14.4 mol)(20.4 J/mol⋅K)+(7.2 mol)(21.1 J/mol⋅S)|=1.31×105 J$

Thus, the initial internal energy of the gas is $1.31×105 J$

Conclusion:

The initial internal energy of the gas is $1.31×105 J$

(c)

To determine

The energy produced if the gas burns out to water vapor completely.

(c)

Expert Solution

Answer to Problem 72AP

The energy produced if the gas burns out to water vapor completely is

$3.48×106 J$ .

Explanation of Solution

Given Info:

The amount of hydrogen in the container is $14.4 mol$ .

The amount of oxygen in the container is $7.2 mol$ .

The initial temperature is $20 °C$

If the gas burns out to water vapor completely, instead of each one mole of hydrogen; one mole water will produce. In the container since the volume is constant, no other work done will be there. The water produced is equal to the amount of hydrogen.

Thus, the amount of water produced is $14.4 mol$ .

The chemical energy required to produce one mole of water is $241.8 kJ$ .

Formula to calculate the energy produced in combustion is,

$ΔU=(241.8 kJ/mol)n$

n is the number of moles of water produced

Substitute $14.4 mol$ for n to find the energy produced,

$ΔU=(241.8 kJ/mol)(14.4 mol)=3.48×106 J$

Thus, the energy produced is $3.48×106 J$ .

Conclusion:

The energy produced if the gas burns out to water vapor completely is $3.48×106 J$ .

(d)

To determine

The temperature and pressure of the steam.

(d)

Expert Solution

Answer to Problem 72AP

The temperature and pressure of the water vapor is

$9.28×103 K$ and

$2.12×106 Pa$ .

Explanation of Solution

Section1:

To determine: The temperature of the steam

Answer: The final temperature of the steam is $9.28×103 K$

Explanation:

Given Info:

The molar heat capacity for water vapor is $27.0 J/mol⋅K$ .

The initial internal energy of the gas is $1.31×105 J$ .

The energy produced if the gas burns out to water vapor completely is $3.48×106 J$

For an ideal gas,

Formula to calculate the temperature of the steam is,

$Tf=UfnwCv,w$ (I)

$Uf$ is the final internal energy
$nw$ is the amount of water in moles
$Cv,w$ is the molar heat capacity of water

Since, after the combustion; the finial internal energy is,

$Uf=Ui+ΔU$

Thus, the equation (I) is given by,

$Tf=Ui+ΔUnwCv,w$

Substitute $1.31×105 J$ for $Ui$ , $3.48×106 J$ for $ΔU$ , $14.4 mol$ for $nw$ and $27.0 J/mol⋅K$ for $Cv,w$ to find the temperature of the steam,

$T=(1.31×105 J)(3.48×106 J)(14.4 mol)(27.0 J/mol⋅K)=9.28×103 K$

Thus, the final temperature of the steam is $9.28×103 K$

Section2:

To determine: The pressure of the steam

Answer: The pressure of the steam is $2.12×106 Pa$

Explanation:

Given Info:

From section1,

The final temperature of the steam is $9.28×103 K$

For an ideal gas,

Formula to calculate the pressure of the steam is,

$P=nwRTV$

$nw$ is the amount of water vapor in moles
R is the universal gas constant
T is the final temperature
V is the volume occupied by the gas

Substitute $14.4 mol$ for $nw$ , $8.31 J/mol⋅K$ for R, $9.28×103 K$ for T and $0.524 m3$ for V to find the pressure of the steam,

$P=(14.4 mol)(8.31 J/mol⋅K)(9.28×103 K)0.524 m3=2.12×106 Pa$

Thus, the pressure of the steam is $2.12×106 Pa$

Conclusion:

The temperature and pressure of the water vapor is $9.28×103 K$ and $2.12×106 Pa$ .

(e)

To determine

The mass of steam and steam’s density.

(e)

Expert Solution

Answer to Problem 72AP

the mass of steam and steams density is

$0.259 kg$ and

$0.494 kg/m3$ .

Explanation of Solution

Section1:

To determine: The mass of steam.

Answer: The mass of the steam is $0.259 kg$ .

Explanation:

Given Info:

The mass of water is $18.0×10−3 kg/mol$

The amount of steam is $14.4 mol$

Formula to calculate the mass of the steam after the combustion is,

$ms=nsMw$

$ns$ is the amount of steam in moles
$Mw$ is the mass of water in moles

Substitute $14.4 mol$ for $ns$ and $18.0×10−3 kg/mol$ for $Mw$ to find the mass of steam,

$ms=(14.4 mol)(18.0×10−3 kg/mol)=0.259 kg$

Thus, mass of the steam is $0.259 kg$ .

Conclusion:

From section1,

Mass of the steam is $0.259 kg$ .

The volume of the gas is $0.524 m3$ .

Formula to calculate the density of the steam is,

$ρs=msV$

$ms$ is the mass of the steam
V is the volume of the gas

Substitute $0.259 kg$ for $ms$ , $0.524 m3$ for V to find the density of the steam,

$ρs=0.259 kg0.524 m3=0.494 kg/m3$

Thus, the steam’s density is $0.494 kg/m3$ .

(f)

To determine

The initial exhaust velocity of the exhaust steam.

(f)

Expert Solution

Answer to Problem 72AP

The initial exhaust velocity of the exhaust steam is

$2.93×103 m/s$ .

Explanation of Solution

Given Info:

The Bernoulli’s equation is given by,

$P2+12ρ2v22+ρsgh2=P1+12ρsv12+ρsgh1$

Assume that the steam in the container is essentially at rest.

Thus,

$v1=0$ , $P2=0$ and $h2=h1$ .

Formula to calculate the exhaust velocity is,

$v2=2P1ρs$

$P1$ is the initial pressure of the gas
$ρs$ is the density of the steam

Substitute $2.12×106 Pa$ for $P1$ and $0.494 kg/m3$ for $ρs$ to find the exhaust velocity,

$v2=2(2.12×106 Pa)0.494 kg/m3=2.93×103 m/s$

Thus, the exhaust velocity is $2.93×103 m/s$ .

Conclusion:

The initial exhaust velocity of the exhaust steam is $2.93×103 m/s$ .

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Answer 3

Question

Chapter 12, Problem 72AP

(a)

To determine

The initial pressure in the vessel.

(a)

Expert Solution

Answer to Problem 72AP

the initial pressure in the vessel is

$1.00×103 Pa$ .

Explanation of Solution

Section1:

To determine: The constant volume occupied by the gas.

Answer: The volume of the gas is $0.524 m3$ .

Explanation:

Given Info:

The radius of the spherical vessel is $0.500 m$ .

Formula to calculate the volume of the gas is,

$V=43πr3$

r is the radius of the vessel

Substitute $0.500 m$ for r to find the volume.

$VSun=43π(0.500 m)3=0.524 m3$

Thus, volume of the gas is $0.524 m3$ .

Conclusion:

From section 1,

The volume of the gas is $0.524 m3$

The amount of hydrogen in the container is $14.4 mol$ .

The amount of oxygen in the container is $7.2 mol$ .

The initial temperature is $20 °C$

To find the initial pressure of the vessel, consider the components in the gas mixture in container as ideal gases. Hence, the total pressure the mixture exerts on the container will be the sum of pressure of the components of the mixture

The pressure exerted by the hydrogen molecule is,

$PH2=nH2RTV$ (I)

$nH2$ is the amount of hydrogen molecules in moles
R is the universal gas constant
T is the absolute temperature
V is the volume of the container

The pressure exerted by the oxygen molecule is,

$PO2=nO2RTV$ (II)

$nO2$ is the amount of oxygen molecule in moles

The total pressure exerted on the container is the sum of equation (I) and (II).

Therefore,

$P=PH2+PO2=(nH2+nO2)RTV$

Substitute $14.4 mol$ for $nH2$ , $7.2 mol$ for $nO2$ , $8.31 J/mol⋅K$ for R, $20 °C$ for T and $0.524 m3$ for V to find the initial pressure of the container,

$P=(14.4 mol+7.2 mol)(8.31 J/mol⋅K)((20+273)K)0.524 m3=1.00×103 Pa$

Thus, the initial pressure in the vessel is $1.00×103 Pa$ .

(b)

To determine

The initial internal energy of the gas.

(b)

Expert Solution

Answer to Problem 72AP

The initial internal energy of the gas is

$1.31×105 J$

Explanation of Solution

Given Info:

The amount of hydrogen in the container is $14.4 mol$ .

The amount of oxygen in the container is $7.2 mol$ .

The molar specific heat of hydrogen at constant volume is $20.4 J/mol.K$ .

The molar specific heat of oxygen at constant volume is $21.1 J/mol⋅K$ .

The internal energy of the gas is the sum of the internal energy of the components of the gas mixture.

Formula to calculate the internal energy of the gas is,

$U=UH2,i+UO2,i=(nO2Cv,o2+nH2Cv,H2)Ti$

$nO2$ is the amount of oxygen in moles
$Cv,o2$ is the molar specific heat of oxygen in constant volume
$nH2$ is the amount of hydrogen in moles
$Cv,H2$ is the molar specific heat of hydrogen in constant volume

Substitute $14.4 mol$ for $nH2$ , $20.4 J/mol.K$ for $Cv,H2$ , $7.2 mol$ $nO2$ and $21.1 J/mol⋅K$ for $Cv,o2$ to find the initial internal energy of the gas,

$Ui=|(14.4 mol)(20.4 J/mol⋅K)+(7.2 mol)(21.1 J/mol⋅S)|=1.31×105 J$

Thus, the initial internal energy of the gas is $1.31×105 J$

Conclusion:

The initial internal energy of the gas is $1.31×105 J$

(c)

To determine

The energy produced if the gas burns out to water vapor completely.

(c)

Expert Solution

Answer to Problem 72AP

The energy produced if the gas burns out to water vapor completely is

$3.48×106 J$ .

Explanation of Solution

Given Info:

The amount of hydrogen in the container is $14.4 mol$ .

The amount of oxygen in the container is $7.2 mol$ .

The initial temperature is $20 °C$

If the gas burns out to water vapor completely, instead of each one mole of hydrogen; one mole water will produce. In the container since the volume is constant, no other work done will be there. The water produced is equal to the amount of hydrogen.

Thus, the amount of water produced is $14.4 mol$ .

The chemical energy required to produce one mole of water is $241.8 kJ$ .

Formula to calculate the energy produced in combustion is,

$ΔU=(241.8 kJ/mol)n$

n is the number of moles of water produced

Substitute $14.4 mol$ for n to find the energy produced,

$ΔU=(241.8 kJ/mol)(14.4 mol)=3.48×106 J$

Thus, the energy produced is $3.48×106 J$ .

Conclusion:

The energy produced if the gas burns out to water vapor completely is $3.48×106 J$ .

(d)

To determine

The temperature and pressure of the steam.

(d)

Expert Solution

Answer to Problem 72AP

The temperature and pressure of the water vapor is

$9.28×103 K$ and

$2.12×106 Pa$ .

Explanation of Solution

Section1:

To determine: The temperature of the steam

Answer: The final temperature of the steam is $9.28×103 K$

Explanation:

Given Info:

The molar heat capacity for water vapor is $27.0 J/mol⋅K$ .

The initial internal energy of the gas is $1.31×105 J$ .

The energy produced if the gas burns out to water vapor completely is $3.48×106 J$

For an ideal gas,

Formula to calculate the temperature of the steam is,

$Tf=UfnwCv,w$ (I)

$Uf$ is the final internal energy
$nw$ is the amount of water in moles
$Cv,w$ is the molar heat capacity of water

Since, after the combustion; the finial internal energy is,

$Uf=Ui+ΔU$

Thus, the equation (I) is given by,

$Tf=Ui+ΔUnwCv,w$

Substitute $1.31×105 J$ for $Ui$ , $3.48×106 J$ for $ΔU$ , $14.4 mol$ for $nw$ and $27.0 J/mol⋅K$ for $Cv,w$ to find the temperature of the steam,

$T=(1.31×105 J)(3.48×106 J)(14.4 mol)(27.0 J/mol⋅K)=9.28×103 K$

Thus, the final temperature of the steam is $9.28×103 K$

Section2:

To determine: The pressure of the steam

Answer: The pressure of the steam is $2.12×106 Pa$

Explanation:

Given Info:

From section1,

The final temperature of the steam is $9.28×103 K$

For an ideal gas,

Formula to calculate the pressure of the steam is,

$P=nwRTV$

$nw$ is the amount of water vapor in moles
R is the universal gas constant
T is the final temperature
V is the volume occupied by the gas

Substitute $14.4 mol$ for $nw$ , $8.31 J/mol⋅K$ for R, $9.28×103 K$ for T and $0.524 m3$ for V to find the pressure of the steam,

$P=(14.4 mol)(8.31 J/mol⋅K)(9.28×103 K)0.524 m3=2.12×106 Pa$

Thus, the pressure of the steam is $2.12×106 Pa$

Conclusion:

The temperature and pressure of the water vapor is $9.28×103 K$ and $2.12×106 Pa$ .

(e)

To determine

The mass of steam and steam’s density.

(e)

Expert Solution

Answer to Problem 72AP

the mass of steam and steams density is

$0.259 kg$ and

$0.494 kg/m3$ .

Explanation of Solution

Section1:

To determine: The mass of steam.

Answer: The mass of the steam is $0.259 kg$ .

Explanation:

Given Info:

The mass of water is $18.0×10−3 kg/mol$

The amount of steam is $14.4 mol$

Formula to calculate the mass of the steam after the combustion is,

$ms=nsMw$

$ns$ is the amount of steam in moles
$Mw$ is the mass of water in moles

Substitute $14.4 mol$ for $ns$ and $18.0×10−3 kg/mol$ for $Mw$ to find the mass of steam,

$ms=(14.4 mol)(18.0×10−3 kg/mol)=0.259 kg$

Thus, mass of the steam is $0.259 kg$ .

Conclusion:

From section1,

Mass of the steam is $0.259 kg$ .

The volume of the gas is $0.524 m3$ .

Formula to calculate the density of the steam is,

$ρs=msV$

$ms$ is the mass of the steam
V is the volume of the gas

Substitute $0.259 kg$ for $ms$ , $0.524 m3$ for V to find the density of the steam,

$ρs=0.259 kg0.524 m3=0.494 kg/m3$

Thus, the steam’s density is $0.494 kg/m3$ .

(f)

To determine

The initial exhaust velocity of the exhaust steam.

(f)

Expert Solution

Answer to Problem 72AP

The initial exhaust velocity of the exhaust steam is

$2.93×103 m/s$ .

Explanation of Solution

Given Info:

The Bernoulli’s equation is given by,

$P2+12ρ2v22+ρsgh2=P1+12ρsv12+ρsgh1$

Assume that the steam in the container is essentially at rest.

Thus,

$v1=0$ , $P2=0$ and $h2=h1$ .

Formula to calculate the exhaust velocity is,

$v2=2P1ρs$

$P1$ is the initial pressure of the gas
$ρs$ is the density of the steam

Substitute $2.12×106 Pa$ for $P1$ and $0.494 kg/m3$ for $ρs$ to find the exhaust velocity,

$v2=2(2.12×106 Pa)0.494 kg/m3=2.93×103 m/s$

Thus, the exhaust velocity is $2.93×103 m/s$ .

Conclusion:

The initial exhaust velocity of the exhaust steam is $2.93×103 m/s$ .

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Answer 4

Question

Chapter 12, Problem 72AP

(a)

To determine

The initial pressure in the vessel.

(a)

Expert Solution

Answer to Problem 72AP

the initial pressure in the vessel is

$1.00×103 Pa$ .

Explanation of Solution

Section1:

To determine: The constant volume occupied by the gas.

Answer: The volume of the gas is $0.524 m3$ .

Explanation:

Given Info:

The radius of the spherical vessel is $0.500 m$ .

Formula to calculate the volume of the gas is,

$V=43πr3$

r is the radius of the vessel

Substitute $0.500 m$ for r to find the volume.

$VSun=43π(0.500 m)3=0.524 m3$

Thus, volume of the gas is $0.524 m3$ .

Conclusion:

From section 1,

The volume of the gas is $0.524 m3$

The amount of hydrogen in the container is $14.4 mol$ .

The amount of oxygen in the container is $7.2 mol$ .

The initial temperature is $20 °C$

To find the initial pressure of the vessel, consider the components in the gas mixture in container as ideal gases. Hence, the total pressure the mixture exerts on the container will be the sum of pressure of the components of the mixture

The pressure exerted by the hydrogen molecule is,

$PH2=nH2RTV$ (I)

$nH2$ is the amount of hydrogen molecules in moles
R is the universal gas constant
T is the absolute temperature
V is the volume of the container

The pressure exerted by the oxygen molecule is,

$PO2=nO2RTV$ (II)

$nO2$ is the amount of oxygen molecule in moles

The total pressure exerted on the container is the sum of equation (I) and (II).

Therefore,

$P=PH2+PO2=(nH2+nO2)RTV$

Substitute $14.4 mol$ for $nH2$ , $7.2 mol$ for $nO2$ , $8.31 J/mol⋅K$ for R, $20 °C$ for T and $0.524 m3$ for V to find the initial pressure of the container,

$P=(14.4 mol+7.2 mol)(8.31 J/mol⋅K)((20+273)K)0.524 m3=1.00×103 Pa$

Thus, the initial pressure in the vessel is $1.00×103 Pa$ .

(b)

To determine

The initial internal energy of the gas.

(b)

Expert Solution

Answer to Problem 72AP

The initial internal energy of the gas is

$1.31×105 J$

Explanation of Solution

Given Info:

The amount of hydrogen in the container is $14.4 mol$ .

The amount of oxygen in the container is $7.2 mol$ .

The molar specific heat of hydrogen at constant volume is $20.4 J/mol.K$ .

The molar specific heat of oxygen at constant volume is $21.1 J/mol⋅K$ .

The internal energy of the gas is the sum of the internal energy of the components of the gas mixture.

Formula to calculate the internal energy of the gas is,

$U=UH2,i+UO2,i=(nO2Cv,o2+nH2Cv,H2)Ti$

$nO2$ is the amount of oxygen in moles
$Cv,o2$ is the molar specific heat of oxygen in constant volume
$nH2$ is the amount of hydrogen in moles
$Cv,H2$ is the molar specific heat of hydrogen in constant volume

Substitute $14.4 mol$ for $nH2$ , $20.4 J/mol.K$ for $Cv,H2$ , $7.2 mol$ $nO2$ and $21.1 J/mol⋅K$ for $Cv,o2$ to find the initial internal energy of the gas,

$Ui=|(14.4 mol)(20.4 J/mol⋅K)+(7.2 mol)(21.1 J/mol⋅S)|=1.31×105 J$

Thus, the initial internal energy of the gas is $1.31×105 J$

Conclusion:

The initial internal energy of the gas is $1.31×105 J$

(c)

To determine

The energy produced if the gas burns out to water vapor completely.

(c)

Expert Solution

Answer to Problem 72AP

The energy produced if the gas burns out to water vapor completely is

$3.48×106 J$ .

Explanation of Solution

Given Info:

The amount of hydrogen in the container is $14.4 mol$ .

The amount of oxygen in the container is $7.2 mol$ .

The initial temperature is $20 °C$

If the gas burns out to water vapor completely, instead of each one mole of hydrogen; one mole water will produce. In the container since the volume is constant, no other work done will be there. The water produced is equal to the amount of hydrogen.

Thus, the amount of water produced is $14.4 mol$ .

The chemical energy required to produce one mole of water is $241.8 kJ$ .

Formula to calculate the energy produced in combustion is,

$ΔU=(241.8 kJ/mol)n$

n is the number of moles of water produced

Substitute $14.4 mol$ for n to find the energy produced,

$ΔU=(241.8 kJ/mol)(14.4 mol)=3.48×106 J$

Thus, the energy produced is $3.48×106 J$ .

Conclusion:

The energy produced if the gas burns out to water vapor completely is $3.48×106 J$ .

(d)

To determine

The temperature and pressure of the steam.

(d)

Expert Solution

Answer to Problem 72AP

The temperature and pressure of the water vapor is

$9.28×103 K$ and

$2.12×106 Pa$ .

Explanation of Solution

Section1:

To determine: The temperature of the steam

Answer: The final temperature of the steam is $9.28×103 K$

Explanation:

Given Info:

The molar heat capacity for water vapor is $27.0 J/mol⋅K$ .

The initial internal energy of the gas is $1.31×105 J$ .

The energy produced if the gas burns out to water vapor completely is $3.48×106 J$

For an ideal gas,

Formula to calculate the temperature of the steam is,

$Tf=UfnwCv,w$ (I)

$Uf$ is the final internal energy
$nw$ is the amount of water in moles
$Cv,w$ is the molar heat capacity of water

Since, after the combustion; the finial internal energy is,

$Uf=Ui+ΔU$

Thus, the equation (I) is given by,

$Tf=Ui+ΔUnwCv,w$

Substitute $1.31×105 J$ for $Ui$ , $3.48×106 J$ for $ΔU$ , $14.4 mol$ for $nw$ and $27.0 J/mol⋅K$ for $Cv,w$ to find the temperature of the steam,

$T=(1.31×105 J)(3.48×106 J)(14.4 mol)(27.0 J/mol⋅K)=9.28×103 K$

Thus, the final temperature of the steam is $9.28×103 K$

Section2:

To determine: The pressure of the steam

Answer: The pressure of the steam is $2.12×106 Pa$

Explanation:

Given Info:

From section1,

The final temperature of the steam is $9.28×103 K$

For an ideal gas,

Formula to calculate the pressure of the steam is,

$P=nwRTV$

$nw$ is the amount of water vapor in moles
R is the universal gas constant
T is the final temperature
V is the volume occupied by the gas

Substitute $14.4 mol$ for $nw$ , $8.31 J/mol⋅K$ for R, $9.28×103 K$ for T and $0.524 m3$ for V to find the pressure of the steam,

$P=(14.4 mol)(8.31 J/mol⋅K)(9.28×103 K)0.524 m3=2.12×106 Pa$

Thus, the pressure of the steam is $2.12×106 Pa$

Conclusion:

The temperature and pressure of the water vapor is $9.28×103 K$ and $2.12×106 Pa$ .

(e)

To determine

The mass of steam and steam’s density.

(e)

Expert Solution

Answer to Problem 72AP

the mass of steam and steams density is

$0.259 kg$ and

$0.494 kg/m3$ .

Explanation of Solution

Section1:

To determine: The mass of steam.

Answer: The mass of the steam is $0.259 kg$ .

Explanation:

Given Info:

The mass of water is $18.0×10−3 kg/mol$

The amount of steam is $14.4 mol$

Formula to calculate the mass of the steam after the combustion is,

$ms=nsMw$

$ns$ is the amount of steam in moles
$Mw$ is the mass of water in moles

Substitute $14.4 mol$ for $ns$ and $18.0×10−3 kg/mol$ for $Mw$ to find the mass of steam,

$ms=(14.4 mol)(18.0×10−3 kg/mol)=0.259 kg$

Thus, mass of the steam is $0.259 kg$ .

Conclusion:

From section1,

Mass of the steam is $0.259 kg$ .

The volume of the gas is $0.524 m3$ .

Formula to calculate the density of the steam is,

$ρs=msV$

$ms$ is the mass of the steam
V is the volume of the gas

Substitute $0.259 kg$ for $ms$ , $0.524 m3$ for V to find the density of the steam,

$ρs=0.259 kg0.524 m3=0.494 kg/m3$

Thus, the steam’s density is $0.494 kg/m3$ .

(f)

To determine

The initial exhaust velocity of the exhaust steam.

(f)

Expert Solution

Answer to Problem 72AP

The initial exhaust velocity of the exhaust steam is

$2.93×103 m/s$ .

Explanation of Solution

Given Info:

The Bernoulli’s equation is given by,

$P2+12ρ2v22+ρsgh2=P1+12ρsv12+ρsgh1$

Assume that the steam in the container is essentially at rest.

Thus,

$v1=0$ , $P2=0$ and $h2=h1$ .

Formula to calculate the exhaust velocity is,

$v2=2P1ρs$

$P1$ is the initial pressure of the gas
$ρs$ is the density of the steam

Substitute $2.12×106 Pa$ for $P1$ and $0.494 kg/m3$ for $ρs$ to find the exhaust velocity,

$v2=2(2.12×106 Pa)0.494 kg/m3=2.93×103 m/s$

Thus, the exhaust velocity is $2.93×103 m/s$ .

Conclusion:

The initial exhaust velocity of the exhaust steam is $2.93×103 m/s$ .

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Answer 5

Chapter 12, Problem 72AP

(a)

To determine

The initial pressure in the vessel.

(a)

Expert Solution

Answer to Problem 72AP

the initial pressure in the vessel is

$1.00×103 Pa$ .

Explanation of Solution

Section1:

To determine: The constant volume occupied by the gas.

Answer: The volume of the gas is $0.524 m3$ .

Explanation:

Given Info:

The radius of the spherical vessel is $0.500 m$ .

Formula to calculate the volume of the gas is,

$V=43πr3$

r is the radius of the vessel

Substitute $0.500 m$ for r to find the volume.

$VSun=43π(0.500 m)3=0.524 m3$

Thus, volume of the gas is $0.524 m3$ .

Conclusion:

From section 1,

The volume of the gas is $0.524 m3$

The amount of hydrogen in the container is $14.4 mol$ .

The amount of oxygen in the container is $7.2 mol$ .

The initial temperature is $20 °C$

To find the initial pressure of the vessel, consider the components in the gas mixture in container as ideal gases. Hence, the total pressure the mixture exerts on the container will be the sum of pressure of the components of the mixture

The pressure exerted by the hydrogen molecule is,

$PH2=nH2RTV$ (I)

$nH2$ is the amount of hydrogen molecules in moles
R is the universal gas constant
T is the absolute temperature
V is the volume of the container

The pressure exerted by the oxygen molecule is,

$PO2=nO2RTV$ (II)

$nO2$ is the amount of oxygen molecule in moles

The total pressure exerted on the container is the sum of equation (I) and (II).

Therefore,

$P=PH2+PO2=(nH2+nO2)RTV$

Substitute $14.4 mol$ for $nH2$ , $7.2 mol$ for $nO2$ , $8.31 J/mol⋅K$ for R, $20 °C$ for T and $0.524 m3$ for V to find the initial pressure of the container,

$P=(14.4 mol+7.2 mol)(8.31 J/mol⋅K)((20+273)K)0.524 m3=1.00×103 Pa$

Thus, the initial pressure in the vessel is $1.00×103 Pa$ .

(b)

To determine

The initial internal energy of the gas.

(b)

Expert Solution

Answer to Problem 72AP

The initial internal energy of the gas is

$1.31×105 J$

Explanation of Solution

Given Info:

The amount of hydrogen in the container is $14.4 mol$ .

The amount of oxygen in the container is $7.2 mol$ .

The molar specific heat of hydrogen at constant volume is $20.4 J/mol.K$ .

The molar specific heat of oxygen at constant volume is $21.1 J/mol⋅K$ .

The internal energy of the gas is the sum of the internal energy of the components of the gas mixture.

Formula to calculate the internal energy of the gas is,

$U=UH2,i+UO2,i=(nO2Cv,o2+nH2Cv,H2)Ti$

$nO2$ is the amount of oxygen in moles
$Cv,o2$ is the molar specific heat of oxygen in constant volume
$nH2$ is the amount of hydrogen in moles
$Cv,H2$ is the molar specific heat of hydrogen in constant volume

Substitute $14.4 mol$ for $nH2$ , $20.4 J/mol.K$ for $Cv,H2$ , $7.2 mol$ $nO2$ and $21.1 J/mol⋅K$ for $Cv,o2$ to find the initial internal energy of the gas,

$Ui=|(14.4 mol)(20.4 J/mol⋅K)+(7.2 mol)(21.1 J/mol⋅S)|=1.31×105 J$

Thus, the initial internal energy of the gas is $1.31×105 J$

Conclusion:

The initial internal energy of the gas is $1.31×105 J$

(c)

To determine

The energy produced if the gas burns out to water vapor completely.

(c)

Expert Solution

Answer to Problem 72AP

The energy produced if the gas burns out to water vapor completely is

$3.48×106 J$ .

Explanation of Solution

Given Info:

The amount of hydrogen in the container is $14.4 mol$ .

The amount of oxygen in the container is $7.2 mol$ .

The initial temperature is $20 °C$

If the gas burns out to water vapor completely, instead of each one mole of hydrogen; one mole water will produce. In the container since the volume is constant, no other work done will be there. The water produced is equal to the amount of hydrogen.

Thus, the amount of water produced is $14.4 mol$ .

The chemical energy required to produce one mole of water is $241.8 kJ$ .

Formula to calculate the energy produced in combustion is,

$ΔU=(241.8 kJ/mol)n$

n is the number of moles of water produced

Substitute $14.4 mol$ for n to find the energy produced,

$ΔU=(241.8 kJ/mol)(14.4 mol)=3.48×106 J$

Thus, the energy produced is $3.48×106 J$ .

Conclusion:

The energy produced if the gas burns out to water vapor completely is $3.48×106 J$ .

(d)

To determine

The temperature and pressure of the steam.

(d)

Expert Solution

Answer to Problem 72AP

The temperature and pressure of the water vapor is

$9.28×103 K$ and

$2.12×106 Pa$ .

Explanation of Solution

Section1:

To determine: The temperature of the steam

Answer: The final temperature of the steam is $9.28×103 K$

Explanation:

Given Info:

The molar heat capacity for water vapor is $27.0 J/mol⋅K$ .

The initial internal energy of the gas is $1.31×105 J$ .

The energy produced if the gas burns out to water vapor completely is $3.48×106 J$

For an ideal gas,

Formula to calculate the temperature of the steam is,

$Tf=UfnwCv,w$ (I)

$Uf$ is the final internal energy
$nw$ is the amount of water in moles
$Cv,w$ is the molar heat capacity of water

Since, after the combustion; the finial internal energy is,

$Uf=Ui+ΔU$

Thus, the equation (I) is given by,

$Tf=Ui+ΔUnwCv,w$

Substitute $1.31×105 J$ for $Ui$ , $3.48×106 J$ for $ΔU$ , $14.4 mol$ for $nw$ and $27.0 J/mol⋅K$ for $Cv,w$ to find the temperature of the steam,

$T=(1.31×105 J)(3.48×106 J)(14.4 mol)(27.0 J/mol⋅K)=9.28×103 K$

Thus, the final temperature of the steam is $9.28×103 K$

Section2:

To determine: The pressure of the steam

Answer: The pressure of the steam is $2.12×106 Pa$

Explanation:

Given Info:

From section1,

The final temperature of the steam is $9.28×103 K$

For an ideal gas,

Formula to calculate the pressure of the steam is,

$P=nwRTV$

$nw$ is the amount of water vapor in moles
R is the universal gas constant
T is the final temperature
V is the volume occupied by the gas

Substitute $14.4 mol$ for $nw$ , $8.31 J/mol⋅K$ for R, $9.28×103 K$ for T and $0.524 m3$ for V to find the pressure of the steam,

$P=(14.4 mol)(8.31 J/mol⋅K)(9.28×103 K)0.524 m3=2.12×106 Pa$

Thus, the pressure of the steam is $2.12×106 Pa$

Conclusion:

The temperature and pressure of the water vapor is $9.28×103 K$ and $2.12×106 Pa$ .

(e)

To determine

The mass of steam and steam’s density.

(e)

Expert Solution

Answer to Problem 72AP

the mass of steam and steams density is

$0.259 kg$ and

$0.494 kg/m3$ .

Explanation of Solution

Section1:

To determine: The mass of steam.

Answer: The mass of the steam is $0.259 kg$ .

Explanation:

Given Info:

The mass of water is $18.0×10−3 kg/mol$

The amount of steam is $14.4 mol$

Formula to calculate the mass of the steam after the combustion is,

$ms=nsMw$

$ns$ is the amount of steam in moles
$Mw$ is the mass of water in moles

Substitute $14.4 mol$ for $ns$ and $18.0×10−3 kg/mol$ for $Mw$ to find the mass of steam,

$ms=(14.4 mol)(18.0×10−3 kg/mol)=0.259 kg$

Thus, mass of the steam is $0.259 kg$ .

Conclusion:

From section1,

Mass of the steam is $0.259 kg$ .

The volume of the gas is $0.524 m3$ .

Formula to calculate the density of the steam is,

$ρs=msV$

$ms$ is the mass of the steam
V is the volume of the gas

Substitute $0.259 kg$ for $ms$ , $0.524 m3$ for V to find the density of the steam,

$ρs=0.259 kg0.524 m3=0.494 kg/m3$

Thus, the steam’s density is $0.494 kg/m3$ .

(f)

To determine

The initial exhaust velocity of the exhaust steam.

(f)

Expert Solution

Answer to Problem 72AP

The initial exhaust velocity of the exhaust steam is

$2.93×103 m/s$ .

Explanation of Solution

Given Info:

The Bernoulli’s equation is given by,

$P2+12ρ2v22+ρsgh2=P1+12ρsv12+ρsgh1$

Assume that the steam in the container is essentially at rest.

Thus,

$v1=0$ , $P2=0$ and $h2=h1$ .

Formula to calculate the exhaust velocity is,

$v2=2P1ρs$

$P1$ is the initial pressure of the gas
$ρs$ is the density of the steam

Substitute $2.12×106 Pa$ for $P1$ and $0.494 kg/m3$ for $ρs$ to find the exhaust velocity,

$v2=2(2.12×106 Pa)0.494 kg/m3=2.93×103 m/s$

Thus, the exhaust velocity is $2.93×103 m/s$ .

Conclusion:

The initial exhaust velocity of the exhaust steam is $2.93×103 m/s$ .

Answer 6

Chapter 12, Problem 72AP

(a)

To determine

The initial pressure in the vessel.

(a)

Expert Solution

Answer to Problem 72AP

the initial pressure in the vessel is

$1.00×103 Pa$ .

Explanation of Solution

Section1:

To determine: The constant volume occupied by the gas.

Answer: The volume of the gas is $0.524 m3$ .

Explanation:

Given Info:

The radius of the spherical vessel is $0.500 m$ .

Formula to calculate the volume of the gas is,

$V=43πr3$

r is the radius of the vessel

Substitute $0.500 m$ for r to find the volume.

$VSun=43π(0.500 m)3=0.524 m3$

Thus, volume of the gas is $0.524 m3$ .

Conclusion:

From section 1,

The volume of the gas is $0.524 m3$

The amount of hydrogen in the container is $14.4 mol$ .

The amount of oxygen in the container is $7.2 mol$ .

The initial temperature is $20 °C$

To find the initial pressure of the vessel, consider the components in the gas mixture in container as ideal gases. Hence, the total pressure the mixture exerts on the container will be the sum of pressure of the components of the mixture

The pressure exerted by the hydrogen molecule is,

$PH2=nH2RTV$ (I)

$nH2$ is the amount of hydrogen molecules in moles
R is the universal gas constant
T is the absolute temperature
V is the volume of the container

The pressure exerted by the oxygen molecule is,

$PO2=nO2RTV$ (II)

$nO2$ is the amount of oxygen molecule in moles

The total pressure exerted on the container is the sum of equation (I) and (II).

Therefore,

$P=PH2+PO2=(nH2+nO2)RTV$

Substitute $14.4 mol$ for $nH2$ , $7.2 mol$ for $nO2$ , $8.31 J/mol⋅K$ for R, $20 °C$ for T and $0.524 m3$ for V to find the initial pressure of the container,

$P=(14.4 mol+7.2 mol)(8.31 J/mol⋅K)((20+273)K)0.524 m3=1.00×103 Pa$

Thus, the initial pressure in the vessel is $1.00×103 Pa$ .

Answer 7

Chapter 12, Problem 72AP

(a)

To determine

The initial pressure in the vessel.

Answer 8

(a)

Expert Solution

Answer to Problem 72AP

the initial pressure in the vessel is

$1.00×103 Pa$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Section1:

To determine: The constant volume occupied by the gas.

Answer: The volume of the gas is $0.524 m3$ .

Explanation:

Given Info:

The radius of the spherical vessel is $0.500 m$ .

Formula to calculate the volume of the gas is,

$V=43πr3$

r is the radius of the vessel

Substitute $0.500 m$ for r to find the volume.

$VSun=43π(0.500 m)3=0.524 m3$

Thus, volume of the gas is $0.524 m3$ .

Conclusion:

From section 1,

The volume of the gas is $0.524 m3$

The amount of hydrogen in the container is $14.4 mol$ .

The amount of oxygen in the container is $7.2 mol$ .

The initial temperature is $20 °C$

To find the initial pressure of the vessel, consider the components in the gas mixture in container as ideal gases. Hence, the total pressure the mixture exerts on the container will be the sum of pressure of the components of the mixture

The pressure exerted by the hydrogen molecule is,

$PH2=nH2RTV$ (I)

$nH2$ is the amount of hydrogen molecules in moles
R is the universal gas constant
T is the absolute temperature
V is the volume of the container

The pressure exerted by the oxygen molecule is,

$PO2=nO2RTV$ (II)

$nO2$ is the amount of oxygen molecule in moles

The total pressure exerted on the container is the sum of equation (I) and (II).

Therefore,

$P=PH2+PO2=(nH2+nO2)RTV$

Substitute $14.4 mol$ for $nH2$ , $7.2 mol$ for $nO2$ , $8.31 J/mol⋅K$ for R, $20 °C$ for T and $0.524 m3$ for V to find the initial pressure of the container,

$P=(14.4 mol+7.2 mol)(8.31 J/mol⋅K)((20+273)K)0.524 m3=1.00×103 Pa$

Thus, the initial pressure in the vessel is $1.00×103 Pa$ .

Answer 13

(b)

To determine

The initial internal energy of the gas.

(b)

Expert Solution

Answer to Problem 72AP

The initial internal energy of the gas is

$1.31×105 J$

Explanation of Solution

Given Info:

The amount of hydrogen in the container is $14.4 mol$ .

The amount of oxygen in the container is $7.2 mol$ .

The molar specific heat of hydrogen at constant volume is $20.4 J/mol.K$ .

The molar specific heat of oxygen at constant volume is $21.1 J/mol⋅K$ .

The internal energy of the gas is the sum of the internal energy of the components of the gas mixture.

Formula to calculate the internal energy of the gas is,

$U=UH2,i+UO2,i=(nO2Cv,o2+nH2Cv,H2)Ti$

$nO2$ is the amount of oxygen in moles
$Cv,o2$ is the molar specific heat of oxygen in constant volume
$nH2$ is the amount of hydrogen in moles
$Cv,H2$ is the molar specific heat of hydrogen in constant volume

Substitute $14.4 mol$ for $nH2$ , $20.4 J/mol.K$ for $Cv,H2$ , $7.2 mol$ $nO2$ and $21.1 J/mol⋅K$ for $Cv,o2$ to find the initial internal energy of the gas,

$Ui=|(14.4 mol)(20.4 J/mol⋅K)+(7.2 mol)(21.1 J/mol⋅S)|=1.31×105 J$

Thus, the initial internal energy of the gas is $1.31×105 J$

Conclusion:

The initial internal energy of the gas is $1.31×105 J$

Answer 14

(b)

To determine

The initial internal energy of the gas.

Answer 15

(b)

Expert Solution

Answer to Problem 72AP

The initial internal energy of the gas is

$1.31×105 J$

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given Info:

The amount of hydrogen in the container is $14.4 mol$ .

The amount of oxygen in the container is $7.2 mol$ .

The molar specific heat of hydrogen at constant volume is $20.4 J/mol.K$ .

The molar specific heat of oxygen at constant volume is $21.1 J/mol⋅K$ .

The internal energy of the gas is the sum of the internal energy of the components of the gas mixture.

Formula to calculate the internal energy of the gas is,

$U=UH2,i+UO2,i=(nO2Cv,o2+nH2Cv,H2)Ti$

$nO2$ is the amount of oxygen in moles
$Cv,o2$ is the molar specific heat of oxygen in constant volume
$nH2$ is the amount of hydrogen in moles
$Cv,H2$ is the molar specific heat of hydrogen in constant volume

Substitute $14.4 mol$ for $nH2$ , $20.4 J/mol.K$ for $Cv,H2$ , $7.2 mol$ $nO2$ and $21.1 J/mol⋅K$ for $Cv,o2$ to find the initial internal energy of the gas,

$Ui=|(14.4 mol)(20.4 J/mol⋅K)+(7.2 mol)(21.1 J/mol⋅S)|=1.31×105 J$

Thus, the initial internal energy of the gas is $1.31×105 J$

Conclusion:

The initial internal energy of the gas is $1.31×105 J$

Answer 20

(c)

To determine

The energy produced if the gas burns out to water vapor completely.

(c)

Expert Solution

Answer to Problem 72AP

The energy produced if the gas burns out to water vapor completely is

$3.48×106 J$ .

Explanation of Solution

Given Info:

The amount of hydrogen in the container is $14.4 mol$ .

The amount of oxygen in the container is $7.2 mol$ .

The initial temperature is $20 °C$

If the gas burns out to water vapor completely, instead of each one mole of hydrogen; one mole water will produce. In the container since the volume is constant, no other work done will be there. The water produced is equal to the amount of hydrogen.

Thus, the amount of water produced is $14.4 mol$ .

The chemical energy required to produce one mole of water is $241.8 kJ$ .

Formula to calculate the energy produced in combustion is,

$ΔU=(241.8 kJ/mol)n$

n is the number of moles of water produced

Substitute $14.4 mol$ for n to find the energy produced,

$ΔU=(241.8 kJ/mol)(14.4 mol)=3.48×106 J$

Thus, the energy produced is $3.48×106 J$ .

Conclusion:

The energy produced if the gas burns out to water vapor completely is $3.48×106 J$ .

Answer 21

(c)

To determine

The energy produced if the gas burns out to water vapor completely.

Answer 22

(c)

Expert Solution

Answer to Problem 72AP

The energy produced if the gas burns out to water vapor completely is

$3.48×106 J$ .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given Info:

The amount of hydrogen in the container is $14.4 mol$ .

The amount of oxygen in the container is $7.2 mol$ .

The initial temperature is $20 °C$

If the gas burns out to water vapor completely, instead of each one mole of hydrogen; one mole water will produce. In the container since the volume is constant, no other work done will be there. The water produced is equal to the amount of hydrogen.

Thus, the amount of water produced is $14.4 mol$ .

The chemical energy required to produce one mole of water is $241.8 kJ$ .

Formula to calculate the energy produced in combustion is,

$ΔU=(241.8 kJ/mol)n$

n is the number of moles of water produced

Substitute $14.4 mol$ for n to find the energy produced,

$ΔU=(241.8 kJ/mol)(14.4 mol)=3.48×106 J$

Thus, the energy produced is $3.48×106 J$ .

Conclusion:

The energy produced if the gas burns out to water vapor completely is $3.48×106 J$ .

Answer 27

(d)

To determine

The temperature and pressure of the steam.

(d)

Expert Solution

Answer to Problem 72AP

The temperature and pressure of the water vapor is

$9.28×103 K$ and

$2.12×106 Pa$ .

Explanation of Solution

Section1:

To determine: The temperature of the steam

Answer: The final temperature of the steam is $9.28×103 K$

Explanation:

Given Info:

The molar heat capacity for water vapor is $27.0 J/mol⋅K$ .

The initial internal energy of the gas is $1.31×105 J$ .

The energy produced if the gas burns out to water vapor completely is $3.48×106 J$

For an ideal gas,

Formula to calculate the temperature of the steam is,

$Tf=UfnwCv,w$ (I)

$Uf$ is the final internal energy
$nw$ is the amount of water in moles
$Cv,w$ is the molar heat capacity of water

Since, after the combustion; the finial internal energy is,

$Uf=Ui+ΔU$

Thus, the equation (I) is given by,

$Tf=Ui+ΔUnwCv,w$

Substitute $1.31×105 J$ for $Ui$ , $3.48×106 J$ for $ΔU$ , $14.4 mol$ for $nw$ and $27.0 J/mol⋅K$ for $Cv,w$ to find the temperature of the steam,

$T=(1.31×105 J)(3.48×106 J)(14.4 mol)(27.0 J/mol⋅K)=9.28×103 K$

Thus, the final temperature of the steam is $9.28×103 K$

Section2:

To determine: The pressure of the steam

Answer: The pressure of the steam is $2.12×106 Pa$

Explanation:

Given Info:

From section1,

The final temperature of the steam is $9.28×103 K$

For an ideal gas,

Formula to calculate the pressure of the steam is,

$P=nwRTV$

$nw$ is the amount of water vapor in moles
R is the universal gas constant
T is the final temperature
V is the volume occupied by the gas

Substitute $14.4 mol$ for $nw$ , $8.31 J/mol⋅K$ for R, $9.28×103 K$ for T and $0.524 m3$ for V to find the pressure of the steam,

$P=(14.4 mol)(8.31 J/mol⋅K)(9.28×103 K)0.524 m3=2.12×106 Pa$

Thus, the pressure of the steam is $2.12×106 Pa$

Conclusion:

The temperature and pressure of the water vapor is $9.28×103 K$ and $2.12×106 Pa$ .

Answer 28

(d)

To determine

The temperature and pressure of the steam.

Answer 29

(d)

Expert Solution

Answer to Problem 72AP

The temperature and pressure of the water vapor is

$9.28×103 K$ and

$2.12×106 Pa$ .

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Section1:

To determine: The temperature of the steam

Answer: The final temperature of the steam is $9.28×103 K$

Explanation:

Given Info:

The molar heat capacity for water vapor is $27.0 J/mol⋅K$ .

The initial internal energy of the gas is $1.31×105 J$ .

The energy produced if the gas burns out to water vapor completely is $3.48×106 J$

For an ideal gas,

Formula to calculate the temperature of the steam is,

$Tf=UfnwCv,w$ (I)

$Uf$ is the final internal energy
$nw$ is the amount of water in moles
$Cv,w$ is the molar heat capacity of water

Since, after the combustion; the finial internal energy is,

$Uf=Ui+ΔU$

Thus, the equation (I) is given by,

$Tf=Ui+ΔUnwCv,w$

Substitute $1.31×105 J$ for $Ui$ , $3.48×106 J$ for $ΔU$ , $14.4 mol$ for $nw$ and $27.0 J/mol⋅K$ for $Cv,w$ to find the temperature of the steam,

$T=(1.31×105 J)(3.48×106 J)(14.4 mol)(27.0 J/mol⋅K)=9.28×103 K$

Thus, the final temperature of the steam is $9.28×103 K$

Section2:

To determine: The pressure of the steam

Answer: The pressure of the steam is $2.12×106 Pa$

Explanation:

Given Info:

From section1,

The final temperature of the steam is $9.28×103 K$

For an ideal gas,

Formula to calculate the pressure of the steam is,

$P=nwRTV$

$nw$ is the amount of water vapor in moles
R is the universal gas constant
T is the final temperature
V is the volume occupied by the gas

Substitute $14.4 mol$ for $nw$ , $8.31 J/mol⋅K$ for R, $9.28×103 K$ for T and $0.524 m3$ for V to find the pressure of the steam,

$P=(14.4 mol)(8.31 J/mol⋅K)(9.28×103 K)0.524 m3=2.12×106 Pa$

Thus, the pressure of the steam is $2.12×106 Pa$

Conclusion:

The temperature and pressure of the water vapor is $9.28×103 K$ and $2.12×106 Pa$ .

Answer 34

(e)

To determine

The mass of steam and steam’s density.

(e)

Expert Solution

Answer to Problem 72AP

the mass of steam and steams density is

$0.259 kg$ and

$0.494 kg/m3$ .

Explanation of Solution

Section1:

To determine: The mass of steam.

Answer: The mass of the steam is $0.259 kg$ .

Explanation:

Given Info:

The mass of water is $18.0×10−3 kg/mol$

The amount of steam is $14.4 mol$

Formula to calculate the mass of the steam after the combustion is,

$ms=nsMw$

$ns$ is the amount of steam in moles
$Mw$ is the mass of water in moles

Substitute $14.4 mol$ for $ns$ and $18.0×10−3 kg/mol$ for $Mw$ to find the mass of steam,

$ms=(14.4 mol)(18.0×10−3 kg/mol)=0.259 kg$

Thus, mass of the steam is $0.259 kg$ .

Conclusion:

From section1,

Mass of the steam is $0.259 kg$ .

The volume of the gas is $0.524 m3$ .

Formula to calculate the density of the steam is,

$ρs=msV$

$ms$ is the mass of the steam
V is the volume of the gas

Substitute $0.259 kg$ for $ms$ , $0.524 m3$ for V to find the density of the steam,

$ρs=0.259 kg0.524 m3=0.494 kg/m3$

Thus, the steam’s density is $0.494 kg/m3$ .

Answer 35

(e)

To determine

The mass of steam and steam’s density.

Answer 36

(e)

Expert Solution

Answer to Problem 72AP

the mass of steam and steams density is

$0.259 kg$ and

$0.494 kg/m3$ .

Answer 37

(e)

Expert Solution

Answer 38

(e)

Expert Solution

Answer 39

Expert Solution

Answer 40

Explanation of Solution

Section1:

To determine: The mass of steam.

Answer: The mass of the steam is $0.259 kg$ .

Explanation:

Given Info:

The mass of water is $18.0×10−3 kg/mol$

The amount of steam is $14.4 mol$

Formula to calculate the mass of the steam after the combustion is,

$ms=nsMw$

$ns$ is the amount of steam in moles
$Mw$ is the mass of water in moles

Substitute $14.4 mol$ for $ns$ and $18.0×10−3 kg/mol$ for $Mw$ to find the mass of steam,

$ms=(14.4 mol)(18.0×10−3 kg/mol)=0.259 kg$

Thus, mass of the steam is $0.259 kg$ .

Conclusion:

From section1,

Mass of the steam is $0.259 kg$ .

The volume of the gas is $0.524 m3$ .

Formula to calculate the density of the steam is,

$ρs=msV$

$ms$ is the mass of the steam
V is the volume of the gas

Substitute $0.259 kg$ for $ms$ , $0.524 m3$ for V to find the density of the steam,

$ρs=0.259 kg0.524 m3=0.494 kg/m3$

Thus, the steam’s density is $0.494 kg/m3$ .

Answer 41

(f)

To determine

The initial exhaust velocity of the exhaust steam.

(f)

Expert Solution

Answer to Problem 72AP

The initial exhaust velocity of the exhaust steam is

$2.93×103 m/s$ .

Explanation of Solution

Given Info:

The Bernoulli’s equation is given by,

$P2+12ρ2v22+ρsgh2=P1+12ρsv12+ρsgh1$

Assume that the steam in the container is essentially at rest.

Thus,

$v1=0$ , $P2=0$ and $h2=h1$ .

Formula to calculate the exhaust velocity is,

$v2=2P1ρs$

$P1$ is the initial pressure of the gas
$ρs$ is the density of the steam

Substitute $2.12×106 Pa$ for $P1$ and $0.494 kg/m3$ for $ρs$ to find the exhaust velocity,

$v2=2(2.12×106 Pa)0.494 kg/m3=2.93×103 m/s$

Thus, the exhaust velocity is $2.93×103 m/s$ .

Conclusion:

The initial exhaust velocity of the exhaust steam is $2.93×103 m/s$ .

Answer 42

(f)

To determine

The initial exhaust velocity of the exhaust steam.

Answer 43

(f)

Expert Solution

Answer to Problem 72AP

The initial exhaust velocity of the exhaust steam is

$2.93×103 m/s$ .

Answer 44

(f)

Expert Solution

Answer 45

(f)

Expert Solution

Answer 46

Expert Solution

Answer 47

Explanation of Solution

Given Info:

The Bernoulli’s equation is given by,

$P2+12ρ2v22+ρsgh2=P1+12ρsv12+ρsgh1$

Assume that the steam in the container is essentially at rest.

Thus,

$v1=0$ , $P2=0$ and $h2=h1$ .

Formula to calculate the exhaust velocity is,

$v2=2P1ρs$

$P1$ is the initial pressure of the gas
$ρs$ is the density of the steam

Substitute $2.12×106 Pa$ for $P1$ and $0.494 kg/m3$ for $ρs$ to find the exhaust velocity,

$v2=2(2.12×106 Pa)0.494 kg/m3=2.93×103 m/s$

Thus, the exhaust velocity is $2.93×103 m/s$ .

Conclusion:

The initial exhaust velocity of the exhaust steam is $2.93×103 m/s$ .

Answer 48

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The initial pressure in the vessel.

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Explanation of Solution

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Answer to Problem 72AP

Explanation of Solution

The energy produced if the gas burns out to water vapor completely.

Answer to Problem 72AP

Explanation of Solution

The temperature and pressure of the steam.

Answer to Problem 72AP

Explanation of Solution

The mass of steam and steam’s density.

Answer to Problem 72AP

Explanation of Solution

The initial exhaust velocity of the exhaust steam.

Answer to Problem 72AP

Explanation of Solution

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