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:** \( RT(2 - \sqrt{2}) \)

Answer 2

Step 1: Recall the formula for work done in an adiabatic process
For an adiabatic process, the work done by the gas when it expands from an initial volume \(V_1\) to a final volume \(V_2\) is given by:
\[ W = \frac{P_1V_1 - P_2V_2}{\gamma - 1} \] where \(P_1, V_1\) and \(P_2, V_2\) are the initial and final pressures and volumes respectively, and \(\gamma\) is the adiabatic constant.

Step 2: Use the adiabatic relation
For an adiabatic process, pressure and volume are related as:
\[ P_1V_1^{\gamma} = P_2V_2^{\gamma} \] This allows us to express \(P_2\) in terms of \(P_1\), \(V_1\), and \(V_2\):
\[ P_2 = P_1 \left( \frac{V_1}{V_2} \right)^{\gamma} \]

Step 3: Substitute in the work expression
Substitute \(P_2\) into the formula for \(W\):
\[ W = \frac{P_1V_1 - P_1V_1 \left( \frac{V_1}{V_2} \right)^{\gamma - 1}}{\gamma - 1} \] Simplify:
\[ W = \frac{P_1V_1}{\gamma - 1} \left[ 1 - \left( \frac{V_1}{V_2} \right)^{\gamma - 1} \right] \]

Step 4: Substitute given values
The problem gives \(V_2 = 2V_1\), \(\gamma = \frac{3}{2}\), and the gas is one mole (\(\mu = 1\)). For one mole of gas, the ideal gas law gives \(P_1V_1 = RT\).
Substitute these into the formula:
\[ W = \frac{RT}{\frac{3}{2} - 1} \left[ 1 - \left( \frac{1}{2} \right)^{\frac{3}{2} - 1} \right] \] Simplify each term:
\[ \frac{3}{2} - 1 = \frac{1}{2} \] and \[ \left( \frac{1}{2} \right)^{\frac{1}{2}} = \frac{1}{\sqrt{2}} \] Substitute back:
\[ W = \frac{RT}{\frac{1}{2}} \left[ 1 - \frac{1}{\sqrt{2}} \right] = 2RT \left( 1 - \frac{1}{\sqrt{2}} \right) \] Simplify the bracket:
\[ W = RT \left[ 2 - \sqrt{2} \right] \]

Step 5: Physical interpretation
During adiabatic expansion, the gas does work on its surroundings by using its internal energy. The negative change in internal energy equals the work done, so the temperature of the gas decreases as it expands. The expression \(RT[2 - \sqrt{2}]\) represents this work done for one mole of gas when the volume doubles under adiabatic conditions.

Final Answer:

\( RT\left[2 - \sqrt{2}\right] \)