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One mole of a monoatomic ideal gas undergoes the process A ⟶ B , as in the given P - V diagram. The specific heat for this process is

Options

  1. A3 R 2
  2. B15 R 7
  3. C30 R 7
  4. D20 R 7

Correct answer

B. 15 R 7

Step-by-step solution

Δ Q = Δ U + W (From 1 s t law) Here, Δ U = n C V Δ T Temperature at A , T A = 2 P 0 V 0 R At B , T B = 16 P 0 V 0 R So, Δ U = 3 2 R 14 P 0 V 0 R = 21 P 0 V 0 W = 2 P 0 3 V 0 + 1 2 3 V 0 2 P 0 [Area under P - V diagram] = 9 P 0 V 0 So, Δ Q = 21 P 0 V 0 + 9 P 0 V 0 = 30 P 0 V 0 Δ Q = n C Δ T C = Δ Q n Δ T = 30 P 0 V 0 14 P 0 V 0 R = 15 7 R

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