Chapter 5, Problem 78QAP

Interpretation Introduction

(a)

Interpretation:

The tank with greater number of moles 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 78QAP

Number of moles of CO₂ and H₂ are same.

Explanation of Solution

The gas 1 is taken as CO₂ and gas 2 as H₂ . For both the gases pressure, volume and temperatures are equal. These values are substituted in combined gas la to determine moles of CO₂ and H₂ as below:

$\begin{array}{l}\stackrel{{\text{CO}}_{\text{2}}}{\overbrace{\frac{\text{PV}}{{\text{n}}_{\text{1}}\text{T}}}}\text{=}\stackrel{{\text{H}}_{\text{2}}}{\overbrace{\frac{\text{PV}}{{\text{n}}_{\text{2}}\text{T}}}}\\ {\text{n}}_{\text{1}}{\text{= n}}_{\text{2}}\end{array}$

Thus, the number of moles of CO₂ and H₂ are same.

Interpretation Introduction

(b)

Interpretation:

The gas 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 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 78QAP

Density of CO₂ is higher than H₂.

Explanation of Solution

The molar mas of CO₂ = 44 g/mol and H₂ = 2g/mol

Thus, MM₁ = 44 g/mol, and MM₂ = 2 g/mol

The ratio of densities of CO₂ and H₂ is determined as below:

$\begin{array}{l}\frac{{\text{d}}_{\text{1}}}{{\text{d}}_{\text{2}}}\text{=}\left(\frac{{\text{MM}}_{\text{1}}\text{P}}{\text{RT}}\right)\left(\frac{\text{RT}}{{\text{MM}}_{\text{2}}\text{P}}\right)\\ \frac{{\text{d}}_{\text{1}}}{{\text{d}}_{\text{2}}}\text{=}\frac{{\text{MM}}_{\text{1}}}{{\text{MM}}_{\text{2}}}\\ \frac{{\text{d}}_{\text{1}}}{{\text{d}}_{\text{2}}}\text{=}\frac{\text{44 g/mol}}{\text{2 g/mol}}\\ {\text{d}}_{\text{1}}{\text{=22(d}}_{\text{2}}\text{)}\end{array}$

Therefore, the density of CO₂ is higher compare to the density of H₂.

Interpretation Introduction

(c)

Interpretation:

The gas with higher effusion time 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 78QAP

Time of effusion of CO₂ will be higher than H₂.

Explanation of Solution

The molar mas of CO₂ = 44 g/mol and H₂ = 2g/mol

The rate of effusion is dependent on the molar mass. The effusion rate for gas with high molar mass will be less. Thus, time of effusion of CO₂ will be higher than that of H₂.

Interpretation Introduction

(d)

Interpretation:

The gas with large 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 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 78QAP

For both the gases average translational energies are same.

Explanation of Solution

The only variable in the equation is temperature. The temperatures for both the gases, CO₂ and H₂ are same. Th ratio of average translation energy of CO₂ and H₂ is calculated as below:

$\begin{array}{l}\frac{{\text{E}}_{\text{t(1)}}}{{\text{E}}_{\text{t(2)}}}\text{=}\left(\frac{\text{3RT}}{{\text{2N}}_{\text{A}}}\right)\left(\frac{{\text{2N}}_{\text{A}}}{\text{3RT}}\right)\\ {\text{E}}_{\text{t(1)}}={\text{E}}_{\text{t(2)}}\end{array}$

Thus, for both the gases CO₂ and H₂ the average translational energies are same.

Interpretation Introduction

(e)

Interpretation:

The gas with more partial pressure on addition of 1 mole of helium in each of the tanks 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 78QAP

Partial pressure for CO₂ and H₂ is same.

Explanation of Solution

The partial pressure of the gas is the pressure exerted by gas alone. As per the ideal gas equation:

$P_1=\frac{n_{1}RT}{V}$

Here partial pressure of gas 1 is P₁, n₁ is the number of moles of gas 1, V is the volume, T is the temperature and R is the universal gas constant

The number of moles of CO₂ and H₂ are same. Thus, the number of moles in tank after adding equal amount of He moles to both the tanks containing CO₂ and H₂ will be same. Also, volume as well as temperature are same of both the gases.

The ratio of partial pressure of tank having CO₂ (i.e. P₁ ) and H₂ (i.e. P₂ ) is determined as below:

$\begin{array}{l}\frac{{\text{P}}_{\text{1}}}{{\text{P}}_{\text{2}}}\text{=}\left(\frac{\text{nrT}}{\text{V}}\right)\left(\frac{\text{V}}{\text{nrT}}\right)\\ {\text{P}}_{\text{1}}={\text{P}}_{\text{2}}\end{array}$

Thus, partial pressure for CO₂ and H₂ are same.