Question

Starting with same initial conditions an ideal gas expands from volume \(V_A\) to \(V_B\) in three different ways. The work done by the gas is 'W\(_1\)' if the process is purely adiabatic, 'W\(_2\)' if it is purely isobaric and 'W\(_3\)' if purely isothermal. Then

• A. \(W_1>W_2>W_3\)
• B. \(W_1<W_2<W_3\)
• C. \(W_2>W_3>W_1\)
• D. \(W_2<W_3<W_1\)

Hint:

Compare the areas under the P-V curves for isobaric, isothermal and adiabatic expansions.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Work done by the gas equals the area under the P-V curve.

Step 2: Order of curves
Starting from the same point, the isobaric line stays highest, the isothermal curve falls, and the adiabatic curve falls most steeply.

Step 3: Compare the areas
Hence the area, and the work, is largest for the isobaric process and smallest for the adiabatic process.

Step 4: Result
\(W_2>W_3>W_1\).

Final Answer:
Isobaric work is greatest, then isothermal, then adiabatic. \[ \boxed{\text{(C) }W_2>W_3>W_1} \]