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

Answer: (D)

23.1 J

Answer 2

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 $R = 8.3$:
$W = 48 - 3 \times (8.3)$
$W = 23.1 \, \text{Joule}$