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An ideal gas with a fixed 2 moles is taken through the process shown on a P-V diagram. The diagram has four key states: A at volume 8 L and pressure 2 atm, B at volume 8 L and pressure P_B, C at volume 10 L and pressure P_C, and D at volume 10 L and pressure 2 atm. The process from B to C is isothermal with temperature T_B = 300 K. Points A and D both lie on the horizontal line P = 2 atm. Which of the following correctly gives the temperature at A and the pressure at B?

1. Temperature at A = 16/R K; Pressure at B = 60R atm
2. Temperature at A = 10/R K; Pressure at B = 75R atm
3. Temperature at A = 8/R K; Pressure at B = 75R atm
4. Temperature at A = 5/R K; Pressure at B = 60R atm

Correct answer: Temperature at A = 8/R K; Pressure at B = 75R atm

Solution

At A: PV = nRT_A → 2 * 8 = 2 * R * T_A → T_A = 8/R (where R is in L*atm*mol⁻¹*K⁻¹). At B: T_B = 300 K, V_B = 8 L, n = 2. P_B * 8 = 2 * R * 300 = 600R → P_B = 75R (in atm, with R in L*atm*mol⁻¹*K⁻¹). This matches option C.

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