One mole of an ideal gas at 27° is taken from A to B as shown in the given PV indicator diagram. The work done by the system will be _________ × 10⁻¹ J. [Given : R = 8.3 J/mole K, ln2 = 0.6931] (Round off to the nearest integer)

Solution

Physics sir · 3 steps

PV is the same at A and B, so this is an isothermal curve: use nRT ln(V₂/V₁).

1. Spot the process. At A, P₁V₁ = 200 × 2 = 400. At B, P₂V₂ = 100 × 4 = 400. Since PV stays constant along the curve, the temperature stays constant too, so the process is isothermal.

   W = nRT ln V₂/V₁

2. Put in numbers. n = 1, T = 27 + 273 = 300 K and V₂/V₁ = 4/2 = 2. So W = 1 × 8.3 × 300 × 0.6931 = 2490 × 0.6931 = 1725.8 J.
3. So. W = 1725.8 J = 17258 × 10⁻¹ J, so the answer is 17258.

The common slip: 1726. Stopping at the work in joules gives 1726. The blank is in units of 10⁻¹ J, so multiply by 10 to get 17258.

The idea to keep. When a curve on a PV diagram has the same product PV at both ends, treat it as isothermal. The work done is then nRT ln(V₂/V₁), and the gas does positive work when it expands.

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