Two moles of an ideal diatomic gas ( $C_{V} = \frac{5}{2} R$ ) are heated from $300 K$ to $340 K$ along the reversible path $P V^{3} = constant$ . Using $R = 8.314 J mo l^{- 1} K^{- 1}$ , find the heat absorbed by the gas in joules (nearest integer).

Numerical answer type, answer in J, tolerance 1.

Answer: 1330 J

Per mole, along

P V^{x} = const

the work done by the gas for a temperature change

Δ T

w_{b y} = - R Δ T / (x - 1)

, so

q = C_{V} Δ T + w_{b y} = (C_{V} + R / (1 - x)) Δ T

. With

x = 3

C = 2.5 R - 0.5 R = 2 R = 16.628 J mo l^{- 1} K^{- 1}

q = 2 \times 16.628 \times 40 = 1330 J