Question

A sample of 1 mole gas at temperature T is adiabatically expanded to double its volume. If adiabatic constant for the gas is \(\gamma=\frac{3}{2}\) , then the work done by the gas in the process is:

• A. \(RT[2-\sqrt2]\)
• B. \(\frac{R}{T}[2-\sqrt2]\)
• C. \(RT[2+\sqrt2]\)
• D. \(\frac{T}{R}[2-\sqrt2]\)

Answer 1

The Correct Option is A

The work performed by the gas during an adiabatic expansion is calculated using the principle of adiabatic processes, where \( PV^\gamma = \text{constant} \). The adiabatic constant \( \gamma \) is given as \( \frac{3}{2} \).

We consider the initial state with pressure \( P \) and volume \( V \), and the final state with pressure \( P' \) and volume \( 2V \), as the volume doubles.

Applying the adiabatic condition \( PV^\gamma = P'(2V)^\gamma \), we derive \( P = P' \cdot 2^\gamma \).

Substituting \( \gamma = \frac{3}{2} \), the relationship becomes \( P = P' \cdot 2^{\frac{3}{2}} \).

The work done \( W \) in an adiabatic process is calculated using the formula \( W = \frac{P_1V_1 - P_2V_2}{\gamma - 1} \).

Substituting the values and rearranging, we get \( W = \frac{P \cdot V - P \cdot 2^{-\frac{1}{2}} \cdot 2V}{\frac{3}{2} - 1} \).

Simplifying this expression yields \( W = \frac{P \cdot V (1 - \sqrt{2})}{\frac{1}{2}} = 2P \cdot V (1 - \sqrt{2}) \).

Using the ideal gas law, \( PV = nRT \), with \( n = 1 \) mole, the work done is \( W = 2 \cdot (RT) \cdot (1 - \sqrt{2}) \).

The final expression for the work done by the gas is \( W = RT[2 - \sqrt{2}] \).

Therefore, the correct option is: \(RT[2 - \sqrt{2}]\).