Chapter 15, Problem 30CR

Interpretation Introduction

(a)

Interpretation:

The volume occupied by $1.15\text{g}$ of helium gas at $25°\text{C}$ and $1.01\text{atm}$ is to be calculated.

Concept Introduction:

The ideal gas equation is used to represent the relation between the volume, pressure, temperature and number of moles of an ideal gas. The ideal gas equation is given as,

$\text{P}\text{V}=\text{n}\text{R}\text{T}$

Where,

• $\text{V}$ represents the volume occupied by the ideal gas.
• $\text{P}$ represents the pressure of the ideal gas.
• $\text{n}$ represents the number of moles of the ideal gas.
• $\text{T}$ represents the temperature of the ideal gas.
• $\text{R}$ represents the ideal gas constant with value $0.08206\text{L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$.

Answer to Problem 30CR

The volume occupied by $1.15\text{g}$ of helium gas at $25°\text{C}$ and $1.01\text{atm}$ is $6.96\text{L}$.

Explanation of Solution

The molar mass of $\text{He}$ gas is $4.00\text{g}\cdot {\text{mol}}^{-1}$.

The mass of $\text{He}$ gas is $1.15\text{g}$.

The temperature of $\text{He}$ gas is $25°\text{C}$.

The temperature of $\text{He}$ gas in Kelvin is given as,

$\begin{array}{c}\text{T}=\left(25°\text{C}+273.15\right)\text{K}\\ =298.15\text{K}\end{array}$

The pressure of $\text{He}$ gas is $1.01\text{atm}$.

The number of moles of substance is given as,

$\text{n}=\frac{\text{m}}{{\text{M}}_{\text{m}}}$

Where,

• $\text{m}$ represents the mass of the substance.
• ${\text{M}}_{\text{m}}$ represents the molar mass of the substance.

Substitute the value of the mass and molar mass of $\text{He}$ gas in the above equation.

$\begin{array}{c}\text{n}=\frac{1.15\text{g}}{4.00\text{g}\cdot {\text{mol}}^{-1}}\\ =0.2875\text{mol}\end{array}$

The number of moles of $\text{He}$ gas is $0.2875\text{mol}$.

The ideal gas equation is given as,

$\text{P}\text{V}=\text{n}\text{R}\text{T}$

Where,

• $\text{V}$ represents the volume occupied by the ideal gas.
• $\text{P}$ represents the pressure of the ideal gas.
• $\text{n}$ represents the number of moles of the ideal gas.
• $\text{T}$ represents the temperature of the ideal gas.
• $\text{R}$ represents the ideal gas constant with value $0.08206\text{L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$.

Rearrange the above equation for the value of $\text{V}$.

` $\text{V}=\frac{\text{n}\text{R}\text{T}}{\text{P}}$

Substitute the value of $\text{P}$, $\text{n}$, $\text{T}$ and $\text{R}$ in the above equation.

$\begin{array}{c}\text{V}=\frac{\left(0.2875\text{mol}\right)\left(0.08206\text{L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(298.15\text{K}\right)}{\left(1.01\text{atm}\right)}\\ =6.96\text{L}\end{array}$

Therefore, the volume occupied by $1.15\text{g}$ of helium gas at $25°\text{C}$ and $1.01\text{atm}$ is $6.96\text{L}$.

Interpretation Introduction

(b)

Interpretation:

The partial pressure of $2.27\text{g}$ of ${\text{H}}_2$ and $1.03\text{g}$ of $\text{He}$ at given temperature and volume is to be calculated.

Concept Introduction:

The ideal gas equation is used to represent the relation between the volume, pressure, temperature and number of moles of an ideal gas. The ideal gas equation is given as,

$\text{P}\text{V}=\text{n}\text{R}\text{T}$

Where,

• $\text{V}$ represents the volume occupied by the ideal gas.
• $\text{P}$ represents the pressure of the ideal gas.
• $\text{n}$ represents the number of moles of the ideal gas.
• $\text{T}$ represents the temperature of the ideal gas.
• $\text{R}$ represents the ideal gas constant with value $0.08206\text{L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$.

Answer to Problem 30CR

The partial pressure of $2.27\text{g}$ of ${\text{H}}_2$ and $1.03\text{g}$ of $\text{He}$ at given temperature and volume is $5.0881\text{atm}$ and $1.1544\text{atm}$ respectively.

Explanation of Solution

The molar mass of ${\text{H}}_2$ gas is $2.00\text{g}\cdot {\text{mol}}^{-1}$.

The mass of ${\text{H}}_2$ gas is $2.27\text{g}$.

The molar mass of $\text{He}$ gas is $4.00\text{g}\cdot {\text{mol}}^{-1}$.

The mass of $\text{He}$ gas is $1.03\text{g}$.

The temperature of container is $0°\text{C}$.

The temperature of container in Kelvin is given as,

$\begin{array}{c}\text{T}=\left(0°\text{C}+273.15\right)\text{K}\\ =273.15\text{K}\end{array}$

The volume of the container is $5.00\text{L}$.

The number of moles of substance is given as:

$\text{n}=\frac{\text{m}}{{\text{M}}_{\text{m}}}$    (1)

Where,

• $\text{m}$ represents the mass of the substance.
• ${\text{M}}_{\text{m}}$ represents the molar mass of the substance.

Substitute the value of the mass and molar mass of ${\text{H}}_2$ gas in the equation (1).

$\begin{array}{c}\text{n}=\frac{2.27\text{g}}{2.00\text{g}\cdot {\text{mol}}^{-1}}\\ =1.135\text{mol}\end{array}$

The number of moles of ${\text{H}}_2$ gas is $1.135\text{mol}$.

Substitute the value of the mass and molar mass of $\text{He}$ gas in the equation (1).

$\begin{array}{c}\text{n}=\frac{1.03\text{g}}{4.00\text{g}\cdot {\text{mol}}^{-1}}\\ =0.2575\text{mol}\end{array}$

The number of moles of $\text{He}$ gas is $0.2575\text{mol}$.

The ideal gas equation is given as,

$\text{P}\text{V}=\text{n}\text{R}\text{T}$

Where,

• $\text{V}$ represents the volume occupied by the ideal gas.
• $\text{P}$ represents the pressure of the ideal gas.
• $\text{n}$ represents the number of moles of the ideal gas.
• $\text{T}$ represents the temperature of the ideal gas.
• $\text{R}$ represents the ideal gas constant with value $0.08206\text{L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$.

Rearrange the above equation for the value of $\text{P}$.

` $\text{P}=\frac{\text{n}\text{R}\text{T}}{\text{V}}$    (2)

Substitute the value of $\text{V}$, $\text{n}$, $\text{T}$ and $\text{R}$ in the equation (2).

$\begin{array}{c}\text{P}=\frac{\left(1.135\text{mol}\right)\left(0.08206\text{L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(273.15\text{K}\right)}{\left(5\text{L}\right)}\\ =5.0881\text{atm}\end{array}$

Therefore, the partial pressure of $2.27\text{g}$ of ${\text{H}}_2$ at given temperature and volume is $5.0881\text{atm}$.

Substitute the value of $\text{V}$, $\text{n}$, $\text{T}$ and $\text{R}$ in the equation (2).

$\begin{array}{c}\text{P}=\frac{\left(0.2575\text{mol}\right)\left(0.08206\text{L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(273.15\text{K}\right)}{\left(5\text{L}\right)}\\ =1.1544\text{atm}\end{array}$

Therefore, the partial pressure of $1.03\text{g}$ of $\text{He}$ at given temperature and volume is $1.1544\text{atm}$.

Interpretation Introduction

(c)

Interpretation:

The pressure existing in a $9.97\text{L}$ tank containing $42.5\text{g}$ of argon gas at $27°\text{C}$ is to be calculated.

Concept Introduction:

The ideal gas equation is used to represent the relation between the volume, pressure, temperature and number of moles of an ideal gas. The ideal gas equation is given as,

$\text{P}\text{V}=\text{n}\text{R}\text{T}$

Where,

• $\text{V}$ represents the volume occupied by the ideal gas.
• $\text{P}$ represents the pressure of the ideal gas.
• $\text{n}$ represents the number of moles of the ideal gas.
• $\text{T}$ represents the temperature of the ideal gas.
• $\text{R}$ represents the ideal gas constant with value $0.08206\text{L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$.

Answer to Problem 30CR

The pressure existing in a $9.97\text{L}$ tank containing $42.5\text{g}$ of argon gas at $27°\text{C}$ is $2.6281\text{atm}$.

Explanation of Solution

The molar mass of argon gas is $39.95\text{g}\cdot {\text{mol}}^{-1}$.

The mass of argon gas is $42.5\text{g}$.

The temperature of tank is $27°\text{C}$.

The temperature of tank in Kelvin is given as,

$\begin{array}{c}\text{T}=\left(27°\text{C}+273.15\right)\text{K}\\ =300.15\text{K}\end{array}$

The volume of the tank is $9.97\text{L}$.

The number of moles of substance is given as,

$\text{n}=\frac{\text{m}}{{\text{M}}_{\text{m}}}$

Where,

• $\text{m}$ represents the mass of the substance.
• ${\text{M}}_{\text{m}}$ represents the molar mass of the substance.

Substitute the value of the mass and molar mass of argon gas in the above equation.

$\begin{array}{c}\text{n}=\frac{42.5\text{g}}{39.95\text{g}\cdot {\text{mol}}^{-1}}\\ =1.0638\text{mol}\end{array}$

The number of moles of argon gas is $1.0638\text{mol}$.

The ideal gas equation is given as,

$\text{P}\text{V}=\text{n}\text{R}\text{T}$

Where,

• $\text{V}$ represents the volume occupied by the ideal gas.
• $\text{P}$ represents the pressure of the ideal gas.
• $\text{n}$ represents the number of moles of the ideal gas.
• $\text{T}$ represents the temperature of the ideal gas.
• $\text{R}$ represents the ideal gas constant with value $0.08206\text{L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$.

Rearrange the above equation for the value of $\text{P}$.

` $\text{P}=\frac{\text{n}\text{R}\text{T}}{\text{V}}$

Substitute the value of $\text{V}$, $\text{n}$, $\text{T}$ and $\text{R}$ in the above equation.

$\begin{array}{c}\text{P}=\frac{\left(1.0638\text{mol}\right)\left(0.08206\text{L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(300.15\text{K}\right)}{\left(9.97\text{L}\right)}\\ =2.6281\text{atm}\end{array}$

Therefore, the pressure existing in a $9.97\text{L}$ tank containing $42.5\text{g}$ of argon gas at $27°\text{C}$ is $2.6281\text{atm}$.