Chapter 13.1, Problem 16SSC

To determine

To calculate:The number of moles and mass of air in the refrigerator.

Answer to Problem 16SSC

  $\text{28}\text{.1}\text{mol}$ ; $\text{0}\text{.815}\text{kg}$

Explanation of Solution

Given data:

Volume of air, V = $\text{0}\text{.635}{\text{m}}^{\text{3}}$

Gas constant, $\text{R=}\text{8}\text{.31}\text{Pa}\cdot {\text{m}}^{\text{3}}\text{/mol}\cdot \text{K}$

Temperature of air, T $\text{=2}°\text{C=275K}$

Molar mass of air, M = $\text{29}\text{g/mol}$

Formula used:

A force (per unit area) that acts on any object is defined by pressure. The ideal gas law is:

  $PV=nRT$

Or, $V=\frac{nRT}{P}$

Or, $n=\frac{PV}{RT}$

Where, n = the number of moles

R = constant, the universal gas constant

P = Pressure and V = Volume

Molar mass of gas:

  $M=\frac{m}{n}$

Mass of gas, $m=nM$

Where, m = mass of gas, M = mass of gas, and n  = number of moles of gas;

Calculation:

It is known that $\text{1}\text{.00}\text{atm}\text{=}\text{1}{\text{.013×10}}^{\text{5}}\text{Pa}$ .

By using the formula $n=\frac{PV}{RT}$ , calculate the number of moles:

  $\begin{array}{c}n=\frac{PV}{RT}\\ \text{=}\frac{\left(\text{1}{\text{.013×10}}^{\text{5}}\text{Pa}\right)\left(\text{0}\text{.635}{\text{m}}^{\text{3}}\right)}{\left(\text{8}\text{.31}\text{Pa}\cdot {\text{m}}^{\text{3}}\text{/mol}\cdot \text{K}\right)\left(\text{275}\right)}\\ \text{=28}\text{.1}\text{mol}\end{array}$

By using the formula $m=nM$ , calculate the mass of gas:

  $\begin{array}{l}m=nM\\ =\left(\text{28}\text{.1}\text{mol}\right)\left(\text{29}\text{g/mol}\right)\\ =814.9\text{g}\\ \text{=}\text{0}\text{.815}\text{kg}\end{array}$

Conclusion:

The number of moles of gas in refrigerator is $\text{28}\text{.1}\text{mol}$ .

The mass of the air refrigerator is $\text{0}\text{.815}\text{kg}$ .