Chapter 13, Problem 33P

(a)

To determine

To Calculate: The volume of nitrogen gas containing tank by using he ideal gas equation.

Answer to Problem 33P

The volume of nitrogen gas containing tank is $14.8{\text{m}}^{\text{3}}$ .

Explanation of Solution

Given:

Mass of nitrogen gas $m=18.5\text{kg}$ .

Formula used:

According to the ideal gas equation.

  $PV=nRT$

  $V=\frac{nRT}{P}$

Where,

  $P=$ Pressure of the gas.

  $V=$ volume of the gas.

  $R=$ Gas constant.

  $T=$ Temperature of the gas.

Calculation:

Consider that the nitrogen gas is the ideal gas.

By using the atomic weight of nitrogen, now calculate the number of moles of the nitrogen atom.

  $\begin{array}{l}n=(18.5\text{kg})\left(\frac{1\text{mol}{\text{N}}_{\text{2}}}{28.0\times {10}^{-3}\text{kg}}\right)=660.71\text{mol}\\ PV=nRT\\ V=\frac{nRT}{P}\\ V=\frac{(660.71\text{mol})(8.315\text{J/mol·K})(273)}{1.013\times {10}^5\text{Pa}}\\ V=14.806{\text{m}}^{\text{3}}\\ V\approx 14.8{\text{m}}^{\text{3}}\end{array}$

Conclusion:

The volume of the nitrogen gas is $14.8{\text{m}}^{\text{3}}$ .

(b)

To determine

To calculate: The change in the pressure if additional nitrogen gas is added without changing the temperature.

Answer to Problem 33P

The change in the pressure is $1.82\text{atm}$ .

Explanation of Solution

Given:

Mass of nitrogen $m=18.5\text{kg }+15\text{kg}$ .

Calculation:

Calculate the number of molecules.

  $\begin{array}{l}n=(18.5\text{kg}+15.0\text{kg})\frac{1\text{mol}{\text{N}}_{\text{2}}}{28.0\times {10}^{-3}\text{kg}}\\ =1196\text{mol}\end{array}$

Write the ideal gas equation.

  $\begin{array}{l}PV=nRT\to P=\frac{nRT}{V}\\ =\frac{(1196\text{mol})(8.315\text{J/mol·K})(273\text{K})}{14.806{\text{m}}^{\text{3}}}\\ =1.83\times {10}^5\text{Pa}\\ =1.82\text{atm}\end{array}$

Conclusion:

The pressure of the nitrogen gas is $1.82\text{atm}$ .