Chapter 13, Problem 32P

To determine

To Calculate: The pressure of the ${\text{CO}}_{\text{2}}$ which is replaced by the nitrogen gas.

Answer to Problem 32P

  $2.32\text{atm}$

Explanation of Solution

Given:

Pressure of nitrogen gas, $P_1=3.65\text{atm}$ .

Mass of the nitrogen gas, $m_1=21.6\text{kg}$ .

Formula used:

Ideal gas equation:

  $PV=nRT$

Where, $P=$ pressure of the gas.

  $T=$ Temperature of the gas.

  $V=$ Volume of the gas.

  $R=$ Gas constant.

  $\begin{array}{l}\frac{P}{n}=\frac{RT}{V}\\ \frac{P_1}{n_1}=\frac{P_2}{n_2}\\ P_2=P_1\frac{n_2}{n_1}\end{array}$

Calculation:

Consider that the nitrogen and ${\text{CO}}_{\text{2}}$ are ideal gas. Volume and temperature of these two gases are constant.

Substituting the values in the above equation

  $\begin{array}{l}=(3.65\text{atm})\left(\frac{\frac{21.6\text{kg}{\text{CO}}_{\text{2}}}{44\times {10}^{-3}\text{kg}{\text{CO}}_{\text{2}}\text{/mol}}}{\frac{21.6\text{kg}{\text{N}}_{\text{2}}}{28\times {10}^{-3}\text{kg}{\text{N}}_{\text{2}}\text{/mol}}}\right)\\ =(3.65\text{atm})\left(\frac{28}{44}\right)\\ =2.32\text{atm}\end{array}$

Conclusion:

Thus, the pressure of the ${\text{CO}}_{\text{2}}$ gas is which is replace by the nitrogen gas is $2.32\text{atm}$ .