A sample of gas with γ = 1.5 is taken through an adiabatic process in which the volume is compressed from 1200 cm³ to 300 cm³. If the initial pressure is 200 kPa. The absolute value of the workdone by the gas in the process = _____________ J.

Solution

Physics sir · 4 steps

Find the final pressure from PV^γ first, then the adiabatic work formula finishes it in one line.

1. Final pressure. In an adiabatic process PV^γ stays constant, so P₂ = P₁(V₁/V₂)^γ. Here V₁/V₂ = 1200/300 = 4 and 4^(1.5) = 8, so P₂ = 8 × 200 = 1600 kPa.

   W = (P₁V₁ - P₂V₂)/(γ - 1)

2. Convert to SI. Use pascals and cubic metres so the products come out in joules. P₁V₁ = 200 × 10³ × 1200 × 10⁻⁶ = 240 J and P₂V₂ = 1600 × 10³ × 300 × 10⁻⁶ = 480 J.
3. Put in numbers. W = (240 - 480)/(1.5 - 1) = -240/0.5 = -480 J. The minus sign means work is done on the gas, as expected for a compression.
4. So. The absolute value of the work done by the gas is 480 J.

The common slip: 240. Forgetting to divide by γ - 1 = 0.5 leaves |P₁V₁ - P₂V₂| = 240, which is only half the work.

The idea to keep. For an adiabatic process, get the missing pressure from PV^γ = constant, then use W = (P₁V₁ - P₂V₂)/(γ - 1). Keep units in Pa and m³ so PV comes out in joules.

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