Question:
2.9 g of a gas (molar mass \(40\,g\,mol^{-1}\)) at \(T(K)\) occupied the same volume as 0.184 g of dihydrogen at \(17^\circ C\) at the same pressure. The value of \(T(K)\) is:
2.9 g of a gas (molar mass \(40\,g\,mol^{-1}\)) at \(T(K)\) occupied the same volume as 0.184 g of dihydrogen at \(17^\circ C\) at the same pressure. The value of \(T(K)\) is:
Show Hint
For equal pressure and equal volume,
\[
\frac{n_1}{n_2}=\frac{T_2}{T_1}
\]
directly follows from the ideal gas equation.
Updated On: Jun 18, 2026
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The Correct Option is B
Solution and Explanation
Concept:
From ideal gas equation,
\[
PV=nRT
\]
At same pressure and volume,
\[
n_1T_1=n_2T_2
\]
Step 1: Calculate moles of first gas.
\[ n_1=\frac{2.9}{40} \] \[ n_1=0.0725 \]
Step 2: Calculate moles of hydrogen.
\[ n_2=\frac{0.184}{2} \] \[ n_2=0.092 \] Temperature: \[ 17^\circ C=290K \]
Step 3: Apply \(nT=\) constant.
\[ n_1T=n_2(290) \] \[ 0.0725T=0.092\times290 \] \[ T=\frac{26.68}{0.0725} \] \[ T\approx368K \] Therefore, \[ \boxed{368K} \]
Step 1: Calculate moles of first gas.
\[ n_1=\frac{2.9}{40} \] \[ n_1=0.0725 \]
Step 2: Calculate moles of hydrogen.
\[ n_2=\frac{0.184}{2} \] \[ n_2=0.092 \] Temperature: \[ 17^\circ C=290K \]
Step 3: Apply \(nT=\) constant.
\[ n_1T=n_2(290) \] \[ 0.0725T=0.092\times290 \] \[ T=\frac{26.68}{0.0725} \] \[ T\approx368K \] Therefore, \[ \boxed{368K} \]
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