Chapter 12, Problem 68AP

(a)

To determine

The initial pressure in the vessel.

Answer to Problem 68AP

the initial pressure in the vessel is $1.00\times {10}^3\text{Pa}$.

Explanation of Solution

Section1:

To determine: The constant volume occupied by the gas.

Answer: The volume of the gas is $0.524{\text{m}}^{\text{3}}$.

Explanation:

Given Info:

The radius of the spherical vessel is $0.500\text{m}$.

Formula to calculate the volume of the gas is,

$V=\frac{4}{3}\pi r^3$

• r is the radius of the vessel

Substitute $0.500\text{m}$ for r to find the volume.

$\begin{array}{c}{V}_{\text{Sun}}=\frac{4}{3}\pi {\left(0.500\text{m}\right)}^3\\ =0.524{\text{m}}^{\text{3}}\end{array}$

Thus, volume of the gas is $0.524{\text{m}}^{\text{3}}$.

Conclusion:

From section 1,

The volume of the gas is $0.524{\text{m}}^{\text{3}}$

The amount of hydrogen in the container is $14.4\text{mol}$.

The amount of oxygen in the container is $7.2\text{mol}$.

The initial temperature is $20\text{°C}$

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,

$P_{H_2}=\frac{n_{H_2}RT}{V}$ (I)

• $n_{H_2}$ 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,

$P_{O_2}=\frac{n_{O_2}RT}{V}$ (II)

• $n_{O_2}$ is the amount of oxygen molecule in moles

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

Therefore,

$\begin{array}{c}P=P_{H_2}+P_{O_2}\\ =\frac{\left(n_{H_2}+n_{O_2}\right)RT}{V}\end{array}$

Substitute $14.4\text{mol}$ for $n_{H_2}$, $7.2\text{mol}$ for $n_{O_2}$, $8.31\text{J/mol}\cdot \text{K}$ for R, $20\text{°C}$ for T and $0.524{\text{m}}^{\text{3}}$ for V to find the initial pressure of the container,

$\begin{array}{c}P=\frac{\left(14.4\text{mol}+7.2\text{mol}\right)\left(8.31\text{J/mol}\cdot \text{K}\right)\left(\left(20+273\right)\text{K}\right)}{0.524{\text{m}}^{\text{3}}}\\ =1.00\times {10}^3\text{Pa}\end{array}$

Thus, the initial pressure in the vessel is $1.00\times {10}^3\text{Pa}$.

(b)

To determine

The initial internal energy of the gas.

Answer to Problem 68AP

The initial internal energy of the gas is $1.31\times {10}^5\text{J}$

Explanation of Solution

Given Info:

The amount of hydrogen in the container is $14.4\text{mol}$.

The amount of oxygen in the container is $7.2\text{mol}$.

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

The molar specific heat of oxygen at constant volume is $21.1\text{J/mol}\cdot \text{K}$.

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,

$\begin{array}{c}U=U_{H_{2,i}}+U_{O_{2,i}}\\ =\left(n_{O_2}{C}_{v,o_2}+n_{H_2}{C}_{v,H_2}\right)T_i\end{array}$

• $n_{O_2}$ is the amount of oxygen in moles
• $C_{v,o_2}$ is the molar specific heat of oxygen in constant volume
• $n_{H_2}$ is the amount of hydrogen in moles
• $C_{v,H_2}$ is the molar specific heat of hydrogen in constant volume

Substitute $14.4\text{mol}$ for $n_{H_2}$, $20.4\text{J/mol}.\text{K}$ for $C_{v,H_2}$, $7.2\text{mol}$ $n_{O_2}$ and $21.1\text{J/mol}\cdot \text{K}$ for $C_{v,o_2}$ to find the initial internal energy of the gas,

$\begin{array}{c}{U}_i=\left|\left(14.4\text{mol}\right)\left(20.4\text{J/mol}\cdot \text{K}\right)+\left(7.2\text{mol}\right)\left(21.1\text{J/mol}\cdot \text{S}\right)\right|\\ =1.31\times {10}^5\text{J}\end{array}$

Thus, the initial internal energy of the gas is $1.31\times {10}^5\text{J}$

Conclusion:

The initial internal energy of the gas is $1.31\times {10}^5\text{J}$

(c)

To determine

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

Answer to Problem 68AP

The energy produced if the gas burns out to water vapor completely is $3.48\times {10}^6\text{J}$.

Explanation of Solution

Given Info:

The amount of hydrogen in the container is $14.4\text{mol}$.

The amount of oxygen in the container is $7.2\text{mol}$.

The initial temperature is $20\text{°C}$

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 $14.4\text{mol}$.

The chemical energy required to produce one mole of water is $241.8\text{kJ}$.

Formula to calculate the energy produced in combustion is,

$\Delta U=\left(241.8\text{kJ/mol}\right)n$

• n is the number of moles of water produced

Substitute $14.4\text{mol}$ for n to find the energy produced,

$\begin{array}{c}\Delta U=\left(241.8\text{kJ/mol}\right)\left(14.4\text{mol}\right)\\ =3.48\times {10}^6\text{J}\end{array}$

Thus, the energy produced is $3.48\times {10}^6\text{J}$.

Conclusion:

The energy produced if the gas burns out to water vapor completely is $3.48\times {10}^6\text{J}$.

(d)

To determine

The temperature and pressure of the steam.

Answer to Problem 68AP

The temperature and pressure of the water vapor is $9.28\times {10}^3\text{K}$ and $2.12\times {10}^6\text{Pa}$.

Explanation of Solution

Section1:

To determine: The temperature of the steam

Answer: The final temperature of the steam is $9.28\times {10}^3\text{K}$

Explanation:

Given Info:

The molar heat capacity for water vapor is $27.0\text{J/mol}\cdot \text{K}$.

The initial internal energy of the gas is $1.31\times {10}^5\text{J}$.

The energy produced if the gas burns out to water vapor completely is $3.48\times {10}^6\text{J}$

For an ideal gas,

Formula to calculate the temperature of the steam is,

$T_f=\frac{U_f}{n_w{C}_{v,w}}$ (I)

• $U_f$ is the final internal energy
• $n_w$ is the amount of water in moles
• $C_{v,w}$ is the molar heat capacity of water

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

$U_f=U_i+\Delta U$

Thus, the equation (I) is given by,

$T_f=\frac{U_i+\Delta U}{n_w{C}_{v,w}}$

Substitute $1.31\times {10}^5\text{J}$ for $U_i$, $3.48\times {10}^6\text{J}$ for $\Delta U$, $14.4\text{mol}$ for $n_w$ and $27.0\text{J/mol}\cdot \text{K}$ for $C_{v,w}$ to find the temperature of the steam,

$\begin{array}{c}T=\frac{\left(1.31\times {10}^5\text{J}\right)\left(3.48\times {10}^6\text{J}\right)}{\left(14.4\text{mol}\right)\left(27.0\text{J/mol}\cdot \text{K}\right)}\\ =9.28\times {10}^3\text{K}\end{array}$

Thus, the final temperature of the steam is $9.28\times {10}^3\text{K}$

Section2:

To determine: The pressure of the steam

Answer: The pressure of the steam is $2.12\times {10}^6\text{Pa}$

Explanation:

Given Info:

From section1,

The final temperature of the steam is $9.28\times {10}^3\text{K}$

For an ideal gas,

Formula to calculate the pressure of the steam is,

$P=\frac{n_{w}RT}{V}$

• $n_w$ 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 $14.4\text{mol}$ for $n_w$, $8.31\text{J/mol}\cdot \text{K}$ for R, $9.28\times {10}^3\text{K}$ for T and $0.524{\text{m}}^{\text{3}}$ for V to find the pressure of the steam,

$\begin{array}{c}P=\frac{\left(14.4\text{mol}\right)\left(8.31\text{J/mol}\cdot \text{K}\right)\left(9.28\times {10}^3\text{K}\right)}{0.524{\text{m}}^{\text{3}}}\\ =2.12\times {10}^6\text{Pa}\end{array}$

Thus, the pressure of the steam is $2.12\times {10}^6\text{Pa}$

Conclusion:

The temperature and pressure of the water vapor is $9.28\times {10}^3\text{K}$ and $2.12\times {10}^6\text{Pa}$.

(e)

To determine

The mass of steam and steam’s density.

Answer to Problem 68AP

the mass of steam and steams density is $0.259\text{kg}$ and $0.494{\text{kg/m}}^{\text{3}}$.

Explanation of Solution

Section1:

To determine: The mass of steam.

Answer: The mass of the steam is $0.259\text{kg}$.

Explanation:

Given Info:

The mass of water is $18.0\times {10}^{-3}\text{kg/mol}$

The amount of steam is $14.4\text{mol}$

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

$m_s=n_s{M}_w$

• $n_s$ is the amount of steam in moles
• $M_w$ is the mass of water in moles

Substitute $14.4\text{mol}$ for $n_s$ and $18.0\times {10}^{-3}\text{kg/mol}$ for $M_w$ to find the mass of steam,

$\begin{array}{c}{m}_s=\left(14.4\text{mol}\right)\left(18.0\times {10}^{-3}\text{kg/mol}\right)\\ =0.259\text{kg}\end{array}$

Thus, mass of the steam is $0.259\text{kg}$.

Conclusion:

From section1,

Mass of the steam is $0.259\text{kg}$.

The volume of the gas is $0.524{\text{m}}^{\text{3}}$.

Formula to calculate the density of the steam is,

${\rho }_s=\frac{m_s}{V}$

• $m_s$ is the mass of the steam
• V is the volume of the gas

Substitute $0.259\text{kg}$ for $m_s$, $0.524{\text{m}}^{\text{3}}$ for V to find the density of the steam,

$\begin{array}{c}{\rho }_s=\frac{0.259\text{kg}}{0.524{\text{m}}^{\text{3}}}\\ =0.494{\text{kg/m}}^{\text{3}}\end{array}$

Thus, the steam’s density is $0.494{\text{kg/m}}^{\text{3}}$.

(f)

To determine

The initial exhaust velocity of the exhaust steam.

Answer to Problem 68AP

The initial exhaust velocity of the exhaust steam is $2.93\times {10}^3\text{m/s}$.

Explanation of Solution

Given Info:

The Bernoulli’s equation is given by,

$P_2+\frac{1}{2}{\rho }_2{v}_2{}^2+{\rho }_{s}g{h}_2=P_1+\frac{1}{2}{\rho }_s{v}_1{}^2+{\rho }_{s}g{h}_1$

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

Thus,

$v_1=0$, $P_2=0$ and $h_2=h_1$.

Formula to calculate the exhaust velocity is,

$v_2=\sqrt{\frac{2P_1}{{\rho }_s}}$

• $P_1$ is the initial pressure of the gas
• ${\rho }_s$ is the density of the steam

Substitute $2.12\times {10}^6\text{Pa}$ for $P_1$ and $0.494{\text{kg/m}}^{\text{3}}$ for ${\rho }_s$ to find the exhaust velocity,

$\begin{array}{c}{v}_2=\sqrt{\frac{2\left(2.12\times {10}^6\text{Pa}\right)}{0.494{\text{kg/m}}^{\text{3}}}}\\ =2.93\times {10}^3\text{m/s}\end{array}$

Thus, the exhaust velocity is $2.93\times {10}^3\text{m/s}$.

Conclusion:

The initial exhaust velocity of the exhaust steam is $2.93\times {10}^3\text{m/s}$.