Question:
A vessel containing nitrogen gas is supplied a heat of \(498\text{ J}\), so as to raise the temperature of the gas by \(40^\circ\text{C}\) at constant pressure. The mass of nitrogen gas in the vessel is
\[
\text{(Molecular mass of nitrogen }=28\text{ g; Universal gas constant }=8.3\text{ J mol}^{-1}\text{K}^{-1}\text{)}
\]
A vessel containing nitrogen gas is supplied a heat of \(498\text{ J}\), so as to raise the temperature of the gas by \(40^\circ\text{C}\) at constant pressure. The mass of nitrogen gas in the vessel is
\[ \text{(Molecular mass of nitrogen }=28\text{ g; Universal gas constant }=8.3\text{ J mol}^{-1}\text{K}^{-1}\text{)} \]
\[ \text{(Molecular mass of nitrogen }=28\text{ g; Universal gas constant }=8.3\text{ J mol}^{-1}\text{K}^{-1}\text{)} \]
Show Hint
For diatomic gases at ordinary temperatures,
\[
C_P=\frac{7R}{2}
\quad \text{and} \quad
C_V=\frac{5R}{2}.
\]
Updated On: Jun 25, 2026
- \(18\text{ g}\)
- \(12\text{ g}\)
- \(20\text{ g}\)
- \(15\text{ g}\)
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Step 1: Use the heat equation at constant pressure.
For a diatomic gas such as nitrogen, \[ C_P=\frac{7R}{2} \] Heat supplied at constant pressure is \[ Q=nC_P\Delta T \] Given: \[ Q=498\text{ J} \] \[ \Delta T=40^\circ\text{C}=40\text{ K} \] \[ R=8.3\text{ J mol}^{-1}\text{K}^{-1} \] Therefore, \[ 498=n\left(\frac{7\times 8.3}{2}\right)(40) \]
Step 2: Simplify to find number of moles.
\[ \frac{7\times 8.3}{2}=29.05 \] So, \[ 498=n(29.05)(40) \] \[ 498=n(1162) \] \[ n=\frac{498}{1162} \] \[ n\approx 0.428\text{ mol} \]
Step 3: Find the mass of nitrogen gas.
Mass: \[ m=nM \] where molecular mass \[ M=28\text{ g mol}^{-1} \] Thus, \[ m=0.428\times 28 \] \[ m\approx 12\text{ g} \]
Step 4: Final conclusion.
Therefore, the mass of nitrogen gas is \[ \boxed{12\text{ g}} \]
For a diatomic gas such as nitrogen, \[ C_P=\frac{7R}{2} \] Heat supplied at constant pressure is \[ Q=nC_P\Delta T \] Given: \[ Q=498\text{ J} \] \[ \Delta T=40^\circ\text{C}=40\text{ K} \] \[ R=8.3\text{ J mol}^{-1}\text{K}^{-1} \] Therefore, \[ 498=n\left(\frac{7\times 8.3}{2}\right)(40) \]
Step 2: Simplify to find number of moles.
\[ \frac{7\times 8.3}{2}=29.05 \] So, \[ 498=n(29.05)(40) \] \[ 498=n(1162) \] \[ n=\frac{498}{1162} \] \[ n\approx 0.428\text{ mol} \]
Step 3: Find the mass of nitrogen gas.
Mass: \[ m=nM \] where molecular mass \[ M=28\text{ g mol}^{-1} \] Thus, \[ m=0.428\times 28 \] \[ m\approx 12\text{ g} \]
Step 4: Final conclusion.
Therefore, the mass of nitrogen gas is \[ \boxed{12\text{ g}} \]
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