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

The objective is to determine the work performed by one mole of helium when it absorbs 48 J of heat, resulting in a temperature increment of \(2^\circ C\).

The process involves a monoatomic ideal gas. The parameters are:

• Heat supplied (\(Q\)) = 48 J
• Temperature rise (\(\Delta T\)) = \(2^\circ C\)
• Ideal gas constant (\(R\)) = 8.3 J K\(^{-1}\) mol\(^{-1}\)

The first law of thermodynamics is applied:

\(Q = \Delta U + W\)

Definitions:

• \(Q\): Heat input to the system.
• \(\Delta U\): Alteration in internal energy.
• \(W\): Work output by the system.

For a monoatomic ideal gas, the internal energy change (\(\Delta U\)) is calculated as:

\(\Delta U = \frac{3}{2}nR\Delta T\)

Given \(n = 1\) mole, the substitution yields:

\(\Delta U = \frac{3}{2} \times 1 \times 8.3 \times 2 = 24.9 \, \text{J}\)

Substituting \(\Delta U\) into the first law:

\(\Delta U = 24.9 \, \text{J}\)

\(48 = 24.9 + W\)

The work done (\(W\)) is:

\(\Delta U = 24.9 \, \text{J}\)

\(W = 48 - 24.9 = 23.1 \, \text{J}\)

Consequently, the work done by the helium is 23.1 J, aligning with the provided correct option.