The P-V diagram of a diatomic ideal gas system going under cyclic process as shown in figure. The work done during an adiabatic process CD is (use γ = 1.4) :

Solution

Physics sir · 3 steps

Read P and V at C and D off the graph, then use the adiabatic work formula with γ − 1 underneath.

1. Read the graph. At C, P_C = 100 N m⁻² and V_C = 4 m³. At D, P_D = 200 N m⁻² and V_D = 3 m³. The gas goes from C to D, so its volume falls: this is a compression.

   W_(CD) = (P_C V_C - P_D V_D)/(γ - 1)

2. Put in numbers. P_C V_C = 100 × 4 = 400 J and P_D V_D = 200 × 3 = 600 J. So W_(CD) = (400 - 600)/(1.4 - 1) = -200/0.4 = -500 J.
3. So. The work done by the gas in CD is -500 J. The minus sign fits a compression, where the surroundings do work on the gas. The answer is (a).

The idea to keep. For an adiabatic process, work done by the gas is (initial PV − final PV)/(γ − 1). Take initial and final from the arrow's direction, and expect a negative answer whenever the volume shrinks.

Answer from the NTA final answer key. Solution by PrepWiser.