Chapter 6, Problem 6.46E

Interpretation Introduction

(a)

Interpretation:

The number of moles of argon in a sample of 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 6.46E

The number of moles of argon in a sample of gas is $0.013\text{mol}$.

Explanation of Solution

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

The temperature of the gas is $90.0°\text{C}$.

The pressure of gas is $735\text{torr}$.

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 $90.0°\text{C}$ to $\text{K}$.

$\begin{array}{rll}90.0°\text{C}& =& \left(90.0+273\right)\text{K}\\ & =& 363\text{K}\end{array}$

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

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

Convert $735\text{torr}$ to $\text{atm}$.

$\begin{array}{rll}735\text{torr}& =& \frac{735}{760}\text{atm}\\ & =& 0.967\text{atm}\end{array}$

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

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

$\begin{array}{rll}0.967\text{atm}\times 0.4\text{L}& =& n\times 0.082\text{L}\text{atm}{\text{K}}^{-1}{\text{mol}}^{-1}\times 363\text{K}\\ n& =& \frac{0.3868\text{atm}\text{L}}{29.76\text{L}\text{atm}}\\ & =& 0.013\text{mol}\end{array}$

Conclusion

The number of moles of argon in a sample of gas is $0.013\text{mol}$.

Interpretation Introduction

(b)

Interpretation:

The pressure exerted by the hydrogen 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 6.46E

The pressure exerted by the hydrogen gas is $7.64\text{atm}$.

Explanation of Solution

The volume of the gas is $2.60\text{L}$.

The temperature of the gas is $50.0°\text{C}$.

The number of moles of gas is $0.750\text{mol}$.

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 $50.0°\text{C}$ to $\text{K}$.

$\begin{array}{rll}50.0°\text{C}& =& \left(50.0+273\right)\text{K}\\ & =& 323\text{K}\end{array}$

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, volume, temperature and gas constant in equation (2).

$\begin{array}{rll}P\times 2.60\text{L}& =& 0.750\text{mol}\times 0.082\text{L}\text{atm}{\text{K}}^{-1}{\text{mol}}^{-1}\times 323\text{K}\\ P& =& \frac{19.86\text{L}\text{atm}}{2.60\text{L}}\\ & =& 7.64\text{atm}\end{array}$

Conclusion

The pressure exerted by the hydrogen gas is $7.64\text{atm}$.

Interpretation Introduction

(c)

Interpretation:

The volume of a tank of nitrogen 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 6.46E

The volume of a tank of nitrogen is $8.34\text{L}$.

Explanation of Solution

The pressure is $4.32\text{atm}$.

The temperature is $20.0°\text{C}$.

The number of moles is $1.50\text{mol}$.

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 $20.0°\text{C}$ to $\text{K}$.

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

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 (3).

$\begin{array}{rll}4.32\text{atm}\times V& =& 1.50\text{mol}\times 0.082\text{L}\text{atm}{\text{K}}^{-1}{\text{mol}}^{-1}\times 293\text{K}\\ V& =& \frac{36.04\text{atm}\text{L}}{4.32\text{atm}}\\ & =& 8.34\text{L}\end{array}$

Conclusion

The volume of a tank of nitrogen is $8.34\text{L}$.