One mole of a monatomic ideal gas goes from $(P_{0}, V_{0})$ at $300 K$ to $(2 P_{0}, 2 V_{0})$ . Path I: isochoric heating to $2 P_{0}$ , then isobaric expansion to $2 V_{0}$ . Path II: isobaric expansion to $2 V_{0}$ , then isochoric heating to $2 P_{0}$ . Take $R = 8.314 J mo l^{- 1} K^{- 1}$ . Select all correct statements.

Answer: A, B, D

The final state has

P V = 4 P_{0} V_{0}

, so

T = 1200 K

and

Δ U = 1.5 R \times 900 = 11224 J

Δ H = 2.5 R \times 900 = 18706 J

for both (state functions). Work is done only in the isobaric step: path I

2 P_{0} \times V_{0} = 2 R T_{0} = 4988 J

; path II

P_{0} V_{0} = 2494 J

. So

q_{I} = 16212 J > q_{II} = 13718 J

, making the third statement false.