Answer
The molar volume of $N_2$ at STP is 22.42 L/mol $N_2$. This is calculated through the ideal gas law.
The molar volume of He at STP is equal to the molar volume of $N_2$ at STP.
The molar volume of $N_2$ is greater than 22.42 L/mol at 1.000 atm and 25.0 degrees Celsius.
The molar volume of $N_2$ collected over water is less than 22.42 L/mol.
Work Step by Step
$PV=nRT\Rightarrow \frac{V}{n}=\frac{RT}{P}\Rightarrow \frac{V}{n}=\frac{273*0.08206}{1.000}=22.42 \text{ L/mol}$
Because pressure and temperature are equal, the two gases have an equal molar volume.
$\frac{P}{V}=\frac{293*0.08206}{1.000}=24.04>22.42 \text{ L\mol}$
Because $N_2$ is collected over water, part of the 1.000 atm pressure is due to the evaporation of water. Therefore, the partial pressure of $N_2$ is less than 1.000 atm thereby proving the molar volume is less than 22.42 L. On the other hand, if the partial pressure of $N_2$ was 1.000 atm, then the molar volume would be 22.42 L in this scenario.
