Question

A total of 48 J heat is given to one mole of helium kept in a cylinder. The temperature of helium increases by $2^\circ C$. The work done by the gas is: (Given, $R = 8.3 \, \text{J K}^{-1} \, \text{mol}^{-1}$).

• A. 72.9 J
• B. 24.9 J
• C. 48 J
• D. 23.1 J

Answer 1

The Correct Option is D

Using the first law of thermodynamics:
\[\Delta Q = \Delta U + W\]
\[31\]
Given:
\[+48 = n C_V \Delta T + W\]
For helium (a monoatomic gas), \( C_V = \frac{3R}{2} \):
\[48 = (1) \left( \frac{3R}{2} \right) (2) + W\]
Simplifying:
\[W = 48 - 3 \times R\]
Substitute \( R = 8.3 \):
\[W = 48 - 3 \times (8.3)\]
\[W = 23.1 \, \text{Joule}\]