Question

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

• A. \( RT\left[\sqrt{2} - 2\right] \)
• B. \( RT\left[1 - 2\sqrt{2}\right] \)
• C. \( RT\left[2\sqrt{2} - 1\right] \)
• D. \( RT\left[2 - \sqrt{2}\right] \)

Answer 1

The Correct Option is D

For an adiabatic process, the work done \( W \) is given by:
\[ W = \frac{nR\Delta T}{1-\gamma}. \]

1. **Using the Adiabatic Condition:**
Since the process is adiabatic, \( TV^{\gamma-1} = \text{constant} \). Let the initial temperature be \( T \) and the final temperature be \( T_f \) when the volume is doubled. Thus,
\[ TV^{\gamma-1} = T_f(2V)^{\gamma-1}. \] 

2. **Calculate \( T_f \):**
Simplifying, we get:
\[ T_f = T \left(\frac{1}{2}\right)^{\frac{\gamma-1}{\gamma}} = T \left(\frac{1}{2}\right)^{\frac{1}{2}} = \frac{T}{\sqrt{2}}. \] 

3. **Calculate the Work Done:**
Substitute into the work formula:
\[ W = \frac{R(T - T_f)}{1 - \frac{3}{2}} = \frac{R \left( T - \frac{T}{\sqrt{2}} \right)}{-\frac{1}{2}}. \] Simplifying further:
\[ W = 2RT\frac{\left(\sqrt{2} - 1\right)}{\sqrt{2}} = RT(2 - \sqrt{2}). \] **Answer:** \( RT(2 - \sqrt{2}) \)