Question

Sol. [Key Concept: For an ideal gas, T = PV/nR and U depends only on temperature; for reversible isobaric expansion, q = nCpΔT and work on the gas is −PΔV.]

At B, pressure is 3.00 atm and volume is 1000 cm³ = 1.000 L. Using the paper’s rounded gas constant gives

$$\text{TB = }\frac{\text{3.00 × 1.000}}{\text{0.100 × 0.082}}\text{ = 365.85 K}$$

Thus A is correct. B → C doubles the volume at constant pressure, so TC = 2TB and ΔT = TB. Since Cv,m = 3R/2, Cp,m = 5R/2.

$$\text{qBC = nCp,mΔT ≈ 0.100 × }\frac{\text{5}}{\text{2}}\text{ × 8.314 × 365.85 ≈ 760 J}$$

Equivalently qBC = (5/2)PΔV = 7.50 L atm ≈ 760 J. This agrees with B to the precision of the listed value.

$$\text{wBC = -PΔV = -3.00(2.00-1.00) = -3.00 L atm ≈ -304 J}$$

Therefore C is incorrect by approximately a factor of ten. Finally, U = U(T) for a perfect gas, so (∂U/∂V)T = 0 and D is correct. The answer is A, B, D.