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Two moles of a monatomic ideal gas undergo a thermodynamic process in which V³ / T² = constant. If the temperature is raised by 300 K, which of the following statements is correct?

1. work done by the gas is 400 R
2. change in internal energy is 900 R
3. molar heat capacity of the gas for the process is 13/6 R
4. molar heat capacity of the gas for the process is 3/2 R

Correct answer: work done by the gas is 400 R

Solution

From V³/T² = const and ideal gas T = PV/nR, we get P*V^(5/3)... actually V³/T² = const => T² proportional to V³ => T proportional to V^(3/2). Using PV = nRT and T ~ V^(3/2): P ~ V^(1/2), meaning PV^(-1/2) = const which means PVⁿ = const with n = -1/2. Molar heat capacity C = Cv + R/(1-n) = (3/2)R + R/(1-(-1/2)) = (3/2)R + (2/3)R = 13/6 R. Work done by 2 moles = n*R*(delta T)/(n_poly - 1) — but n_poly = -1/2, so W = 2*R*300/(1-(-1/2)) would be negative. Let me recompute: W = nR*delta_T/(1-n_poly) = 2*R*300/(1-(-1/2)) = 600R/(3/2) = 400R. So work done = 400R.

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