Question

A gas is initially at a state of \( (P_0, V_0, T_0) \). If \( W_1 \) is work done in isobaric expansion to volume \( V \) and \( W_2 \) is the work done in isothermal expansion to volume \( V \), then:

• A. \( W_1 = W_2 \)
• B. \( W_1<W_2 \)
• C. \( W_1>W_2 \)
• D. \( W_1 - W_2 = R \)

Hint:

Remember that isothermal processes often involve less work due to the logarithmic nature of the pressure-volume relationship, compared to linear relationships in isobaric processes.

Answer 1

The Correct Option is C

Step 1: Calculate work done in isobaric process. \[ W_1 = P_0(V - V_0) \] Step 2: Calculate work done in isothermal process. \[ W_2 = P_0V_0 \ln\left(\frac{V}{V_0}\right) \] Step 3: Compare \( W_1 \) and \( W_2 \). For most practical ranges of \( V \) and \( V_0 \), \( W_1 \) is greater than \( W_2 \) due to the nature of the logarithmic function in the isothermal process.

Answer 2

To solve this problem, we need to compare the work done in an isobaric expansion and an isothermal expansion for a gas. The question gives us the work done in these two processes and asks for their relationship.

1. Understanding the Two Processes:
- In an isobaric expansion, the work done \( W_1 \) is given by the formula: \[ W_1 = P \Delta V \] where \( P \) is the constant pressure and \( \Delta V \) is the change in volume. - In an isothermal expansion, the work done \( W_2 \) is given by the formula: \[ W_2 = nRT \ln \frac{V}{V_0} \] where \( n \) is the number of moles, \( R \) is the universal gas constant, \( T \) is the temperature, and \( V \) and \( V_0 \) are the final and initial volumes, respectively. In the isobaric process, the work done depends on the constant pressure and the change in volume. In contrast, for the isothermal process, the work done depends on the logarithmic change in volume, which generally results in a larger amount of work compared to the isobaric process for the same change in volume.

2. Conclusion:
Since the work done in an isothermal expansion is generally greater than the work done in an isobaric expansion for the same volume change, we have: \[ W_1 < W_2 \]

Final Answer:
The correct answer is Option B: \( W_1 < W_2 \).