One mole of a monatomic gas is taken from a point A to another point N along with the path ACB . The initial temperature at A is T 0 . Calculate the heat absorbed by the gas in the process A → C → B

Step-by-step solution

If T B be the temperature at B, then by gas law P A V A T A = P B V B T B ∴ T B = P B V B T B T A = 2 P 0 2 V 0 P 0 V 0 T 0 The change in internal energy from A to B Δ U = n C v Δ T = 1 × 3 R 2 × 4 T 0 - T 0 = 9 RT 0 2 Work done in the process A to C W AC = P Δ V = P 0 2 V 0 - V 0 = P 0 V 0 = RT 0 and W CB = 0 ∴ Total work done from A → C → B W AC + W CB = RT 0 + 0 = RT 0 From the first law of thermodynamics, Q = Δ U + W = 9 RT 0 2 + RT 0 = 1 1 RT 0 2 Thus heat absorbed by the gas from A → C → B is 1 1 RT 0 2