Chapter 5, Problem 79QAP

Interpretation Introduction

(a)

Interpretation:

The tank containing higher mass needs to be determined.

Concept introduction:

According to the ideal gas law volume i.e. V, pressure i.e. P, number of moles i.e. m, temperature i.e. t and universal gas constant i.e. R are interrelated as below:

$\text{PV = nRT}$

When at two different conditions gases are placed, then to determine the changed variable combined gas law is used. Below is the formula of combined gas law:

$\stackrel{\text{Gas 1}}{\overbrace{\frac{{\text{P}}_{\text{1}}{\text{V}}_{\text{1}}}{{\text{n}}_{\text{1}}{\text{T}}_{\text{1}}}}}\text{=}\stackrel{\text{Gas2}}{\overbrace{\frac{{\text{P}}_{\text{2}}{\text{V}}_{\text{2}}}{{\text{n}}_{\text{2}}{\text{T}}_{\text{2}}}}}$

Here

• P₁ and P₂ are the pressure of gases
• V₁ and V₂ and volume of gases
• n₁ and n₂ number of moles
• T₁ and T₂ are the temperature of gases

Gas density is known as the ratio of mass of the gas and the volume occupied by that gas. The formula is as below:

$\begin{array}{l}\text{d = }\frac{\text{m}}{\text{v}}\\ \text{ = }\frac{\left(\text{MM}\right)\text{P}}{\text{RT}}\end{array}$

Here, MM is the molar mass of the gas, V is the volume of gas, P is pressure of gas, T is temperature and R is the universal gas constant.

The kinetic model of gases is accounted for ideal gas behavior. The formula of average translational energy of gas is as below:

${\text{E}}_{\text{t}}\text{=}\frac{\text{3RT}}{{\text{2N}}_{\text{A}}}$

Here,

E_t = average translational energy of gas

T = temperature in Kelvin

R = Universal gas constant

N_A = Avogadro number

Effusion is known as the leakage of gas molecules from high to low pressure region via a pinhole. For any two gas molecules the formula to determine the time needed for effusion is as below:

$\frac{{\text{time}}_{\text{1}}}{{\text{time}}_{\text{2}}}\text{=}{\left(\frac{{\text{MM}}_{\text{1}}}{{\text{MM}}_{\text{2}}}\right)}^{\text{1/2}}$

Here time₁ and time₂ is the time of effusion for gas1 and gas 2. MM₁ and MM₂ is the molar mass for gas1 and gas 2.

Answer to Problem 79QAP

The mass of SO₂ in tank A is GT the mass of O₂ in tank B.

Explanation of Solution

Below is the expression of ideal gas equation:

$\text{PV = nRT }\mathrm{..........}\left(\text{1}\right)$

Here, volume i.e. V, pressure i.e. P, number of moles i.e. m, temperature i.e. t and universal gas constant i.e. R

$n=\frac{PV}{RT}=\text{Constant}$

We have $\text{Molar Mass =}\frac{\text{Mass}}{\text{n}}\mathrm{.........}\text{ (2)}$

Thus, no of moles is constant then molar mass is directly proportional to the mass of gases and if the molar mass is greater then mass of the gas will be greater as well.

So, let’s consider the number of moles in each tank to be one:

$\begin{array}{l}{\text{Molar mass of SO}}_{\text{2}}=32+16\times 2=64\text{ g}\\ {\text{Molar mass of O}}_{\text{2}}=16\times 2=32\text{ g}\end{array}$

Thus, the mass of SO₂ in tank A is GT the mass of O₂ in tank B.

Interpretation Introduction

(b)

Interpretation:

The gas between CH₄ and SO₂ with higher average translational energy needs to be determined.

Concept introduction:

According to the ideal gas law volume i.e. V, pressure i.e. P, number of moles i.e. m, temperature i.e. t and universal gas constant i.e. R are interrelated as below:

$\text{PV = nRT}$

When at two different conditions gases are placed, then to determine the changed variable combined gas law is used. Below is the formula of combined gas law:

$\stackrel{\text{Gas 1}}{\overbrace{\frac{{\text{P}}_{\text{1}}{\text{V}}_{\text{1}}}{{\text{n}}_{\text{1}}{\text{T}}_{\text{1}}}}}\text{=}\stackrel{\text{Gas2}}{\overbrace{\frac{{\text{P}}_{\text{2}}{\text{V}}_{\text{2}}}{{\text{n}}_{\text{2}}{\text{T}}_{\text{2}}}}}$

Here

• P₁ and P₂ are the pressure of gases
• V₁ and V₂ and volume of gases
• n₁ and n₂ number of moles
• T₁ and T₂ are the temperature of gases

Gas density is known as the ratio of mass of the gas and the volume occupied by that gas. The formula is as below:

$\begin{array}{l}\text{d = }\frac{\text{m}}{\text{v}}\\ \text{ = }\frac{\left(\text{MM}\right)\text{P}}{\text{RT}}\end{array}$

Here, MM is the molar mass of the gas, V is the volume of gas, P is pressure of gas, T is temperature and R is the universal gas constant.

The kinetic model of gases is accounted for ideal gas behavior. The formula of average translational energy of gas is as below:

${\text{E}}_{\text{t}}\text{=}\frac{\text{3RT}}{{\text{2N}}_{\text{A}}}$

Here,

E_t = average translational energy of gas

T = temperature in Kelvin

R = Universal gas constant

N_A = Avogadro number

Effusion is known as the leakage of gas molecules from high to low pressure region via a pinhole. For any two gas molecules the formula to determine the time needed for effusion is as below:

$\frac{{\text{time}}_{\text{1}}}{{\text{time}}_{\text{2}}}\text{=}{\left(\frac{{\text{MM}}_{\text{1}}}{{\text{MM}}_{\text{2}}}\right)}^{\text{1/2}}$

Here, time₁ and time₂ is the time of effusion for gas1 and gas 2. MM₁ and MM₂ is the molar mass for gas1 and gas 2.

Answer to Problem 79QAP

The average kinetic average translational energy of CH₄ in tank C is EQ the average kinetic average translational energy of SO₂ in tank A.

Explanation of Solution

Now we know that according to the ideal gas law the value of the average translational kinetic energy is determined as below:

${\text{E}}_{\text{t}}\text{=}\frac{\text{3RT}}{{\text{2N}}_{\text{A}}}$

Here,

E_t = average translational energy of gas

T = temperature in Kelvin

R = Universal gas constant

N_A = Avogadro number

Because, the temperature is also constant. So, both the gases will contain similar average kinetic average translational energy.

Therefore, the average kinetic average translational energy of CH₄ in tank C is EQ the average kinetic average translational energy of SO₂ in tank A.

Interpretation Introduction

(c)

Interpretation:

The effusion time of SO₂ from Tank A needs to be determined.

Concept introduction:

According to the ideal gas law volume i.e. V, pressure i.e. P, number of moles i.e. m, temperature i.e. t and universal gas constant i.e. R is interrelated as below:

$\text{PV = nRT}$

When at two different conditions gases are placed, then to determine the changed variable combined gas law is used. Below is the formula of combined gas law:

$\stackrel{\text{Gas 1}}{\overbrace{\frac{{\text{P}}_{\text{1}}{\text{V}}_{\text{1}}}{{\text{n}}_{\text{1}}{\text{T}}_{\text{1}}}}}\text{=}\stackrel{\text{Gas2}}{\overbrace{\frac{{\text{P}}_{\text{2}}{\text{V}}_{\text{2}}}{{\text{n}}_{\text{2}}{\text{T}}_{\text{2}}}}}$

Here

• P₁ and P₂ are the pressure of gases
• V₁ and V₂ and volume of gases
• n₁ and n₂ number of moles
• T₁ and T₂ are the temperature of gases

Gas density is known as the ratio of mass of the gas and the volume occupied by that gas. The formula is as below:

$\begin{array}{l}\text{d = }\frac{\text{m}}{\text{v}}\\ \text{ = }\frac{\left(\text{MM}\right)\text{P}}{\text{RT}}\end{array}$

Here, MM is the molar mass of the gas, V is the volume of gas, P is pressure of gas, T is temperature and R is the universal gas constant.

The kinetic model of gases is accounted for ideal gas behavior. The formula of average translational energy of gas is as below:

${\text{E}}_{\text{t}}\text{=}\frac{\text{3RT}}{{\text{2N}}_{\text{A}}}$

Here,

E_t = average translational energy of gas

T = temperature in Kelvin

R = Universal gas constant

N_A = Avogadro number

Effusion is known as the leakage of gas molecules from high to low pressure region via a pinhole. For any two gas molecules the formula to determine the time needed for effusion is as below:

$\frac{{\text{time}}_{\text{1}}}{{\text{time}}_{\text{2}}}\text{=}{\left(\frac{{\text{MM}}_{\text{1}}}{{\text{MM}}_{\text{2}}}\right)}^{\text{1/2}}$

Here time₁ and time₂ is the time of effusion for gas1 and gas 2. MM₁ and MM₂ is the molar mass for gas1 and gas 2.

Answer to Problem 79QAP

Time taken by total of SO₂ gas to effuse out of tank from an identical hole is LT 40s.

Explanation of Solution

As per the Graham’s law of effusion (at constant P and T) is as below:

$\frac{{\text{rate of effusion SO}}_{\text{2}}}{{\text{rate of effusion O}}_{\text{2}}}=\sqrt{\frac{{\text{M}}_{{\text{O}}_{\text{2}}}}{{\text{M}}_{{\text{SO}}_{\text{2}}}}}$

Here M_O2 is the molar mass of oxygen gas and M_SO2 is the molar mass of Sulphur dioxide gas

Putting the values:

$\begin{array}{l}\frac{{\text{rate of effusion SO}}_{\text{2}}}{{\text{rate of effusion O}}_{\text{2}}}=\sqrt{\frac{\text{32}}{\text{64}}}=0.707\\ {\text{rate of effusion SO}}_{\text{2}}=0.707\times {\text{rate of effusion O}}_{\text{2}}\end{array}$

Thus, of O₂ gas take 20 Sec to effuse then the time taken by SO₂ gas to effuse:

$\begin{array}{l}\text{t = 0}\text{.707}\times \text{20}\\ \text{ = 14}\text{.14 sec}\end{array}$

Therefore, time taken by total of SO₂ gas to effuse out of tank from an identical hole is LT 40s.

Interpretation Introduction

(d)

Interpretation:

The gas from O₂ and CH_4, with higher density needs to be determined.

Concept introduction:

According to the ideal gas law volume i.e. V, pressure i.e. P, number of moles i.e. m, temperature i.e. t and universal gas constant i.e. R is interrelated as below:

$\text{PV = nRT}$

When at two different conditions gases are placed, then to determine the changed variable combined gas law is used. Below is the formula of combined gas law:

$\stackrel{\text{Gas 1}}{\overbrace{\frac{{\text{P}}_{\text{1}}{\text{V}}_{\text{1}}}{{\text{n}}_{\text{1}}{\text{T}}_{\text{1}}}}}\text{=}\stackrel{\text{Gas2}}{\overbrace{\frac{{\text{P}}_{\text{2}}{\text{V}}_{\text{2}}}{{\text{n}}_{\text{2}}{\text{T}}_{\text{2}}}}}$

Here

• P₁ and P₂ are the pressure of gases
• V₁ and V₂ and volume of gases
• n₁ and n₂ number of moles
• T₁ and T₂ are the temperature of gases

Gas density is known as the ratio of mass of the gas and the volume occupied by that gas. The formula is as below:

$\begin{array}{l}\text{d = }\frac{\text{m}}{\text{v}}\\ \text{ = }\frac{\left(\text{MM}\right)\text{P}}{\text{RT}}\end{array}$

Here, MM is the molar mass of the gas, V is the volume of gas, P is pressure of gas, T is temperature and R is the universal gas constant.

The kinetic model of gases is accounted for ideal gas behavior. The formula of average translational energy of gas is as below:

${\text{E}}_{\text{t}}\text{=}\frac{\text{3RT}}{{\text{2N}}_{\text{A}}}$

Here,

E_t = average translational energy of gas

T = temperature in Kelvin

R = Universal gas constant

N_A = Avogadro number

Effusion is known as the leakage of gas molecules from high to low pressure region via a pinhole. For any two gas molecules the formula to determine the time needed for effusion is as below:

$\frac{{\text{time}}_{\text{1}}}{{\text{time}}_{\text{2}}}\text{=}{\left(\frac{{\text{MM}}_{\text{1}}}{{\text{MM}}_{\text{2}}}\right)}^{\text{1/2}}$

Here time₁ and time₂ is the time of effusion for gas1 and gas 2. MM₁ and MM₂ is the molar mass for gas1 and gas 2.

Answer to Problem 79QAP

For both the gases average translational energies are same.

Explanation of Solution

The expression of the density is as below:

$\text{d = }\frac{\text{M}}{\text{V}}$

Here M is the molar mass and V is volume.

Thus, when volume is constant then density is directly proportional to molar mass so as greater the molar mass greater will be the density:

Now in case of O₂ and CH₄

$\begin{array}{l}{\text{Molar mass of O}}_{\text{2}}=2\times 16=32\text{ g}\\ {\text{Molar mass of CH}}_{\text{4}}=12+1\times 4=16\text{ g}\end{array}$

Thus, molar mass of oxygen is greater than the molar lass of methane gas.

Therefore, the densities of O₂ in tank B is GT the densities of gas CH₄ in tank C.

Interpretation Introduction

(e)

Interpretation:

The tank between A and B with higher pressure needs to be determined.

Concept introduction:

According to the ideal gas law volume i.e. V, pressure i.e. P, number of moles i.e. m, temperature i.e. t and universal gas constant i.e. R is interrelated as below:

$\text{PV = nRT}$

When at two different conditions gases are placed, then to determine the changed variable combined gas law is used. Below is the formula of combined gas law:

$\stackrel{\text{Gas 1}}{\overbrace{\frac{{\text{P}}_{\text{1}}{\text{V}}_{\text{1}}}{{\text{n}}_{\text{1}}{\text{T}}_{\text{1}}}}}\text{=}\stackrel{\text{Gas2}}{\overbrace{\frac{{\text{P}}_{\text{2}}{\text{V}}_{\text{2}}}{{\text{n}}_{\text{2}}{\text{T}}_{\text{2}}}}}$

Here

• P₁ and P₂ are the pressure of gases
• V₁ and V₂ and volume of gases
• n₁ and n₂ number of moles
• T₁ and T₂ are the temperature of gases

Gas density is known as the ratio of mass of the gas and the volume occupied by that gas. The formula is as below:

$\begin{array}{l}\text{d = }\frac{\text{m}}{\text{v}}\\ \text{ = }\frac{\left(\text{MM}\right)\text{P}}{\text{RT}}\end{array}$

Here, MM is the molar mass of the gas, V is the volume of gas, P is pressure of gas, T is temperature and R is the universal gas constant.

The kinetic model of gases is accounted for ideal gas behavior. The formula of average translational energy of gas is as below:

${\text{E}}_{\text{t}}\text{=}\frac{\text{3RT}}{{\text{2N}}_{\text{A}}}$

Here,

E_t = average translational energy of gas

T = temperature in Kelvin

R = Universal gas constant

N_A = Avogadro number

Effusion is known as the leakage of gas molecules from high to low pressure region via a pinhole. For any two gas molecules the formula to determine the time needed for effusion is as below:

$\frac{{\text{time}}_{\text{1}}}{{\text{time}}_{\text{2}}}\text{=}{\left(\frac{{\text{MM}}_{\text{1}}}{{\text{MM}}_{\text{2}}}\right)}^{\text{1/2}}$

Here, time₁ and time₂ is the time of effusion for gas1 and gas 2. MM₁ and MM₂ is the molar mass for gas1 and gas 2.

Answer to Problem 79QAP

For both the gases average translational energies are same.

Explanation of Solution

According to the ideal gas equation:

$\text{PV = nRT}$

$\text{T = }\frac{\text{PV}}{\text{nR}}$

If n, V is constant as in the question, then T is directly proportional to pressure so:

$\frac{{\text{T}}_{\text{A}}}{{\text{T}}_{\text{B}}}=\frac{{\text{P}}_{\text{A}}}{{\text{P}}_{\text{B}}}$

Here, T_A and T_B are the temperatures of Tank A and B while P_A and P_B is the pressure of Tank A and B.

Putting the values of T_A and T_B

$\begin{array}{l}\frac{\text{300K}}{\text{150K}}=\frac{{\text{P}}_{\text{A}}}{{\text{P}}_{\text{B}}}\\ {\text{P}}_{\text{A}}=2{\text{P}}_{\text{B}}\end{array}$

Thus, the pressure in tank A is GT half the pressure in tank B.