Chapter 6, Problem 6.45E

Interpretation Introduction

(a)

Interpretation:

The pressure exerted by $2.00\text{mol}$ of oxygen confined to a volume of $400\text{mL}$ at $20.0°\text{C}$ is to be calculated.

Concept introduction:

According to the ideal gas law, the relation between pressure, temperature, volume and number of moles is,

$PV=nRT$

Answer to Problem 6.45E

The pressure exerted by $2.00\text{mol}$ of oxygen confined to a volume of $400\text{mL}$ at $20.0°\text{C}$ is $120.1\text{atm}$.

Explanation of Solution

The volume of the gas is $400\text{mL}$.

The temperature of the gas is $20.0°\text{C}$.

The number of moles of gas is $2.00\text{mol}$.

According to the ideal gas law, the relation between pressure, temperature, volume and number of moles is,

$PV=nRT$ …(1)

Where,

• $P$ is the pressure of the gas.

• $V$ is the volume of the gas.

• $R$ is the universal gas constant.

• $n$ is the number of moles of gas.

• $T$ is the temperature of the gas.

Convert $20.0°\text{C}$ to $\text{K}$.

$\begin{array}{rll}20.0°\text{C}& =& \left(20.0+273\right)\text{K}\\ & =& 293\text{K}\end{array}$

Convert $400\text{mL}$ to $\text{L}$.

$\begin{array}{rll}400\text{mL}& =& \frac{400}{1000}\text{L}\\ & =& 0.4\text{L}\end{array}$

The value of universal gas constant is $0.082\text{L}\text{atm}{\text{K}}^{-1}{\text{mol}}^{-1}$.

Substitute the value of number of moles, volume, temperature and gas constant in equation (1).

$\begin{array}{rll}P\times 0.4\text{L}& =& 2.00\text{mol}\times 0.082\text{L}\text{atm}{\text{K}}^{-1}{\text{mol}}^{-1}\times 293\text{K}\\ P& =& \frac{48.05\text{L}\text{atm}}{0.4\text{L}}\\ & =& 120.1\text{atm}\end{array}$

Conclusion

The pressure exerted by $2.00\text{mol}$ of oxygen confined to a volume of $400\text{mL}$ at $20.0°\text{C}$ is $120.1\text{atm}$.

Interpretation Introduction

(b)

Interpretation:

The volume of hydrogen gas in a steel cylinder is to be calculated.

Concept introduction:

According to the ideal gas law, the relation between pressure, temperature, volume and number of moles is,

$PV=nRT$

Answer to Problem 6.45E

The volume of hydrogen gas in a steel cylinder is $4.13\text{L}$.

Explanation of Solution

The pressure of the gas is $3.00\text{atm}$.

The temperature of the gas is $15.0°\text{C}$.

The number of moles of gas is $0.525\text{mol}$.

According to the ideal gas law, the relation between pressure, temperature, volume and number of moles is,

$PV=nRT$ …(2)

Where,

• $P$ is the pressure of the gas.

• $V$ is the volume of the gas.

• $R$ is the universal gas constant.

• $n$ is the number of moles of gas.

• $T$ is the temperature of the gas.

Convert $15.0°\text{C}$ to $\text{K}$.

$\begin{array}{rll}15.0°\text{C}& =& \left(15.0+273\right)\text{K}\\ & =& 288\text{K}\end{array}$

The value of universal gas constant is $0.082\text{L}\text{atm}{\text{K}}^{-1}{\text{mol}}^{-1}$.

Substitute the value of number of moles, pressure, temperature and gas constant in equation (2).

$\begin{array}{rll}3.00\text{atm}\times V& =& 0.525\text{mol}\times 0.082\text{L}\text{atm}{\text{K}}^{-1}{\text{mol}}^{-1}\times 288\text{K}\\ V& =& \frac{12.4\text{L}\text{atm}}{3.00\text{atm}}\\ & =& 4.13\text{L}\end{array}$

Conclusion

The volume of hydrogen gas in a steel cylinder is $4.13\text{L}$.

Interpretation Introduction

(c)

Interpretation:

The temperature of a nitrogen gas is to be calculated.

Concept introduction:

According to the ideal gas law, the relation between pressure, temperature, volume and number of moles is,

$PV=nRT$

Answer to Problem 6.45E

The temperature of a nitrogen gas is $108.38\text{K}$.

Explanation of Solution

The pressure of the gas is $300\text{torr}$.

The volume of the gas is $2.25\text{L}$.

The number of moles of gas is $0.100\text{mol}$.

According to the ideal gas law, the relation between pressure, temperature, volume and number of moles is,

$PV=nRT$ …(3)

Where,

• $P$ is the pressure of the gas.

• $V$ is the volume of the gas.

• $R$ is the universal gas constant.

• $n$ is the number of moles of gas.

• $T$ is the temperature of the gas.

Convert $300\text{torr}$ to $\text{atm}$.

$\begin{array}{rll}300\text{torr}& =& \frac{300}{760}\text{atm}\\ & =& 0.395\text{atm}\end{array}$

The value of universal gas constant is $0.082\text{L}\text{atm}{\text{K}}^{-1}{\text{mol}}^{-1}$.

Substitute the value of number of moles, pressure, volume and gas constant in equation (3).

$\begin{array}{rll}0.395\text{atm}\times 2.25\text{L}& =& 0.100\text{mol}\times 0.082\text{L}\text{atm}{\text{K}}^{-1}{\text{mol}}^{-1}\times T\\ T& =& \frac{0.395\text{atm}\times 2.25\text{L}}{0.100\text{mol}\times 0.082\text{L}\text{atm}{\text{K}}^{-1}{\text{mol}}^{-1}}\\ & =& 108.38\text{K}\end{array}$

Conclusion

The temperature of a nitrogen gas is $108.38\text{K}$.