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

To solve the problem, we need to calculate the work done by one mole of helium when given 48 J of heat, causing its temperature to increase by \(2^\circ C\).

First, we understand the process: Helium is a monoatomic ideal gas. The given parameters are:

• Heat added (\(Q\)) = 48 J
• Temperature increase (\(\Delta T\)) = \(2^\circ C\)
• Gas constant (\(R\)) = 8.3 J K\(^{-1}\) mol\(^{-1}\)

We use the first law of thermodynamics, which states:

\(Q = \Delta U + W\)

Where:

• \(Q\) is the heat added to the system.
• \(\Delta U\) is the change in internal energy.
• \(W\) is the work done by the system.

For a monoatomic ideal gas, the change in internal energy (\(\Delta U\)) can be given by:

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

Since we have one mole of helium, \(n = 1\). Substituting the given values, we get:

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

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

\(48 = 24.9 + W\)

Solving for \(W\):

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

Thus, the work done by the gas is 23.1 J. This matches the correct option provided.

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