Chapter 13.1, Problem 9PP

To determine

To Calculate:The volume of 1.0 kg of air at atmospheric pressure.

Answer to Problem 9PP

  $0.8\text{3}{\text{m}}^{\text{3}}$

Explanation of Solution

Given data:

Pressure, $P=1\text{atm}=1.01\times {10}^{\text{5}}\text{Pa}$ (known)

Temperature, $T=20.0°\text{C}+273=293\text{K}$

Value of gas constant, $R=8.31\text{Pa}\cdot {\text{m}}^{\text{3}}\text{/mol}\cdot \text{K}$

The molar mass of gas is $\text{29}\text{g/mol}$ .

Formula used:

From the ideal gas law:

  $PV=nRT$

  $V=\frac{nRT}{P}$

Where, n is the number of moles, P is the pressure, T is the temperature, V is the volume and R is the gas constant.

Molar mass of gas, $M=\frac{m}{n}$

Mass of gas, $m=Mn$

Where, n is the number of moles.

Calculation:

Number of moles of gas,

  $\begin{array}{c}n=\frac{m}{M}\\ =\frac{\text{1}{\text{.0×10}}^{\text{3}}\text{g}}{\text{29}\text{g/mol}}\end{array}$

Now, calculate volume by using ideal gas law:

  $\begin{array}{c}V=\frac{nRT}{P}\\ \text{=}\frac{\left(\frac{\text{1}{\text{.0×10}}^{\text{3}}\text{g}}{\text{29}\text{g/mol}}\right)\left(\text{8}\text{.31}\text{Pa}\cdot {\text{m}}^{\text{3}}\text{/mol}\cdot \text{K}\right)\left(\text{293}\text{K}\right)}{\left(\text{1}{\text{.013×10}}^{\text{5}}\text{Pa}\right)}\\ \text{=0}\text{.83}{\text{m}}^{\text{3}}\end{array}$

Conclusion:

The volume of 1.0 kg of air at atmospheric pressure is $0.8\text{3}{\text{m}}^{\text{3}}$ .