Question:
At $T(K)$ the kinetic energy of one mole of an ideal gas is 3735 J. If its pressure is 1 atm, the volume of gas is ($R = 8.3~J~mol^{-1}K^{-1}$)
At $T(K)$ the kinetic energy of one mole of an ideal gas is 3735 J. If its pressure is 1 atm, the volume of gas is ($R = 8.3~J~mol^{-1}K^{-1}$)
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For one mole ideal gas:
\[
KE=\frac32RT
\]
and
\[
PV=RT
\]
These two formulas are commonly combined in thermodynamics problems.
Updated On: Jun 17, 2026
- $12.3~L$
- $24.6~L$
- $49.2~L$
- $36.8~L$
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The Correct Option is B
Solution and Explanation
Concept:
For one mole of an ideal gas,
\[
KE=\frac{3}{2}RT
\]
Also,
\[
PV=RT
\]
Step 1: Calculate the temperature.
Given, \[ \frac{3}{2}RT=3735 \] \[ T=\frac{2\times 3735}{3\times 8.3} \] \[ T=300K \]
Step 2: Use ideal gas equation.
\[ PV=RT \] At 1 atm, \[ V=\frac{RT}{P} \] \[ V=\frac{8.3\times300}{1.013\times10^5} \] \[ V\approx2.46\times10^{-2}m^3 \] \[ V=24.6L \] \[ \boxed{24.6L} \]
Step 1: Calculate the temperature.
Given, \[ \frac{3}{2}RT=3735 \] \[ T=\frac{2\times 3735}{3\times 8.3} \] \[ T=300K \]
Step 2: Use ideal gas equation.
\[ PV=RT \] At 1 atm, \[ V=\frac{RT}{P} \] \[ V=\frac{8.3\times300}{1.013\times10^5} \] \[ V\approx2.46\times10^{-2}m^3 \] \[ V=24.6L \] \[ \boxed{24.6L} \]
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