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.