Chapter 5, Problem 80QAP

Interpretation Introduction

(a)

Interpretation:

The bulb 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 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:

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

Here u₁ and u₂ is the rate of effusion for gas1 and gas 2. MM₁ and MM₂ is the molar mass for gas1 and gas 2.

The whole pressure (P_tot ) exerted by the combination of gases equal to the addition of the partial pressure exerted through each gas.

As per the Dalton law of partial pressure mole fraction is the mole of each component present in total moles of compound:

${\text{P}}_{\text{1}}\text{=}\left(\frac{{\text{n}}_{\text{1}}}{{\text{n}}_{\text{tot}}}\right){\text{P}}_{\text{tot}}$

Here, n₁ is the mole of component, n_tot is total moles and n₁ /n_tot mole fraction of gas 1.

Answer to Problem 80QAP

Pressure in both the bulbs will be same.

Explanation of Solution

The gas present in bulb A is N₂ and bulb is NH₃ . The moles of N₂ and NH₃ is 1 mol each. In both bulbs, the volume is same, and temperature is also same. The ideal gas equation is used to determine the ratio of pressure in bulb A (P_A ) and bulb B (P_B ) as below:

$\frac{{\text{P}}_{\text{A}}}{{\text{P}}_{\text{B}}}\text{ = }\frac{\left({\text{n}}_{\text{A}}\text{RT}\right)}{\left({\text{n}}_{\text{B}}\text{RT}\right)}$

The ratio of both the gases are same which means n_A and n_B and same. Putting these values in the above equation:

$\begin{array}{l}\frac{{\text{P}}_{\text{A}}}{{\text{P}}_{\text{B}}}\text{ = }\frac{\left(\frac{{\text{n}}_{\text{A}}\text{RT}}{\text{V}}\right)}{\left(\frac{{\text{n}}_{\text{B}}\text{RT}}{\text{V}}\right)}\\ {\text{P}}_{\text{A}}={\text{P}}_{\text{B}}\end{array}$

Thus, the pressure in both the bulbs would be same.

Interpretation Introduction

(b)

Interpretation:

The bulb containing 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 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:

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

Here u₁ and u₂ is the rate of effusion for gas1 and gas 2. MM₁ and MM₂ is the molar mass for gas1 and gas 2.

The whole pressure (P_tot ) exerted by the combination of gases equal to the addition of the partial pressure exerted through each gas.

As per the Dalton law of partial pressure mole fraction is the mole of each component present in total moles of compound:

${\text{P}}_{\text{1}}\text{=}\left(\frac{{\text{n}}_{\text{1}}}{{\text{n}}_{\text{tot}}}\right){\text{P}}_{\text{tot}}$

Here, n₁ is the mole of component, n_tot is total moles and n₁ /n_tot mole fraction of gas 1.

Answer to Problem 80QAP

Bulb A has the higher density.

Explanation of Solution

The gas density is directly proportional to the molar mass of the gas. The molar mass of N₂ is 28 g/ml and NH₃ is 17 g/mol. The molar mass of N₂ is higher than NH₃ . Thus, the density of N₂ would be higher.

So, the bulb A having the higher density.

Interpretation Introduction

(c)

Interpretation:

The bulb containing the molecule having higher kinetic 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:

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

Here u₁ and u₂ is the rate of effusion for gas1 and gas 2. MM₁ and MM₂ is the molar mass for gas1 and gas 2.

The whole pressure (P_tot ) exerted by the combination of gases equal to the addition of the partial pressure exerted through each gas.

As per the Dalton law of partial pressure mole fraction is the mole of each component present in total moles of compound:

${\text{P}}_{\text{1}}\text{=}\left(\frac{{\text{n}}_{\text{1}}}{{\text{n}}_{\text{tot}}}\right){\text{P}}_{\text{tot}}$

Here, n₁ is the mole of component, n_tot is total moles and n₁ /n_tot mole fraction of gas 1.

Answer to Problem 80QAP

For both the bulbs kinetic energy is same.

Explanation of Solution

The average kinetic energy formula displays that average kinetic energy is directly proportional to the kinetic temperature. The temperatures in both the bulbs are same. Thus, the kinetic energy in bulb A and bulb B would be same.

Interpretation Introduction

(d)

Interpretation:

The bulb containing the gas that moves with the faster molecular speed 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:

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

Here u₁ and u₂ is the rate of effusion for gas1 and gas 2. MM₁ and MM₂ is the molar mass for gas1 and gas 2.

The whole pressure (P_tot ) exerted by the combination of gases equal to the addition of the partial pressure exerted through each gas.

As per the Dalton law of partial pressure mole fraction is the mole of each component present in total moles of compound:

${\text{P}}_{\text{1}}\text{=}\left(\frac{{\text{n}}_{\text{1}}}{{\text{n}}_{\text{tot}}}\right){\text{P}}_{\text{tot}}$

Here, n₁ is the mole of component, n_tot is total moles and n₁ /n_tot mole fraction of gas 1.

Answer to Problem 80QAP

NH₃ having the higher average speed.

Explanation of Solution

As per the effusion formula shows the average effusion speed of a gas is inversely proportional to the square root of the molar mass of the gas. This means the gas with higher molar mass will have slower average speed for the effusion. The molar mass of N₂ is higher than NH₃ . Thus, the gas having the higher average speed would be NH₃.

Interpretation Introduction

(e)

Interpretation:

The change in pressure when valve between the two bulbs is opened needs to be calculated.

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 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:

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

Here u₁ and u₂ is the rate of effusion for gas1 and gas 2. MM₁ and MM₂ is the molar mass for gas1 and gas 2.

The whole pressure (P_tot ) exerted by the combination of gases equal to the addition of the partial pressure exerted through each gas.

As per the Dalton law of partial pressure mole fraction is the mole of each component present in total moles of compound:

${\text{P}}_{\text{1}}\text{=}\left(\frac{{\text{n}}_{\text{1}}}{{\text{n}}_{\text{tot}}}\right){\text{P}}_{\text{tot}}$

Here, n₁ is the mole of component, n_tot is total moles and n₁ /n_tot mole fraction of gas 1.

Answer to Problem 80QAP

Pressure in both the bulbs will be same.

Explanation of Solution

The open valve will maintain equilibrium amid gases in each bulb. Because all the conditions like number of moles, volume and temperature are same. Thus, the pressure would be same.

Interpretation Introduction

(f)

Interpretation:

The fraction of the total pressure due to He gas 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:

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

Here u₁ and u₂ is the rate of effusion for gas1 and gas 2. MM₁ and MM₂ is the molar mass for gas1 and gas 2.

The whole pressure (P_tot ) exerted by the combination of gases equal to the addition of the partial pressure exerted through each gas.

As per the Dalton law of partial pressure mole fraction is the mole of each component present in total moles of compound:

${\text{P}}_{\text{1}}\text{=}\left(\frac{{\text{n}}_{\text{1}}}{{\text{n}}_{\text{tot}}}\right){\text{P}}_{\text{tot}}$

Here, n₁ is the mole of component, n_tot is total moles and n₁ /n_tot mole fraction of gas 1.

Answer to Problem 80QAP

The fraction of helium of the total pressure would be ½.

Explanation of Solution

The moles of He added in the system is 2 mol. The moles of N₂ is 1 mol and NH₃ is 1 mol. The total mole number of gases in the system is calculated as below:

$\begin{array}{l}{\text{n}}_{\text{tot}}{\text{= n}}_{{\text{N}}_{\text{2}}}+{\text{ n}}_{{\text{N}}_{{\text{NH}}_3}}+{\text{ n}}_{\text{He}}\\ \text{ = }\left(\text{1 mol}\right)+\left(\text{1 mol}\right)+\left(\text{2 mol}\right)\\ \text{ = 4 mol}\end{array}$

Thus, the total number of moles is 4 mol.

The values of n_tot and n_He is substituted in equation of Dalton’s law of partial pressure shown below:

$\begin{array}{l}{\text{P}}_{\text{He}}\text{=}\left(\frac{{\text{n}}_{\text{He}}}{{\text{n}}_{\text{tot}}}\right){\text{P}}_{\text{tot}}\\ {\text{P}}_{\text{He}}\text{=}\left(\frac{2}{4}\right){\text{P}}_{\text{tot}}\\ \frac{{\text{P}}_{\text{He}}}{{\text{P}}_{\text{tot}}}=\frac{1}{2}\end{array}$

Thus, the fraction of helium to the total pressure would be $\frac{1}{2}$.