A fixed mass of a gas is first heated isobarically to double the volume and then cooled isochorically to decrease the temperature back to the initial value. By what factor would the work done by the decreased, had the process been isothermal?

Options

A. 2
B. `1/2`
C. ln 2
D. ln 3

Answer

To solve the problem, we need to analyze the two processes: the isobaric process followed by the isochoric process, and then compare it to an isothermal process.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions:**
- Let the initial volume of the gas be \( V_0 \) and the initial pressure be \( P_0 \).
- The initial temperature is \( T_0 \).

2. **Isobaric Heating to Double the Volume:**
- The gas is heated isobarically (at constant pressure) to double its volume. Therefore, the final volume after this process becomes:
\[ V_f = 2V_0 \]
- Since the process is isobaric, the work done \( W_1 \) during this process can be calculated using the formula:
\[ W_1 = P \Delta V = P_0 (V_f - V_0) = P_0 (2V_0 - V_0) = P_0 V_0 \]

3. **Isochoric Cooling to Initial Temperature:**
- After reaching \( 2V_0 \), the gas is cooled isochorically (at constant volume) to return to the initial temperature \( T_0 \).
- During this isochoric process, no work is done since the volume remains constant:
\[ W_2 = 0 \]

4. **Total Work Done in the Given Process:**
- The total work done in the entire process (isobaric followed by isochoric) is:
\[ W_{\text{total}} = W_1 + W_2 = P_0 V_0 + 0 = P_0 V_0 \]

5. **Isothermal Process:**
- Now, let's consider the hypothetical scenario where the gas undergoes an isothermal process from the initial state \( (V_0, T_0) \) to the final state \( (2V_0, T_0) \).
- The work done \( W_{\text{isothermal}} \) in an isothermal process is given by:
\[ W_{\text{isothermal}} = nRT \ln\left(\frac{V_f}{V_i}\right) \]
- Here, since \( nRT = P_0 V_0 \) at the initial state, we can express the work done as:
\[ W_{\text{isothermal}} = P_0 V_0 \ln\left(\frac{2V_0}{V_0}\right) = P_0 V_0 \ln(2) \]

6. **Comparing Work Done:**
- Now, we need to find the factor by which the work done in the original process decreases compared to the isothermal process:
\[ \text{Factor} = \frac{W_{\text{isothermal}}}{W_{\text{total}}} = \frac{P_0 V_0 \ln(2)}{P_0 V_0} = \ln(2) \]

### Final Answer:
The factor by which the work done decreases if the process had been isothermal is:
\[ \text{Factor} = \ln(2) \]