One mole of a monatomic ideal gas at $2.0 atm$ and $10 L$ is expanded at constant pressure to $20 L$ , and then cooled at constant volume until the pressure is $1.0 atm$ . Taking $1 L atm = 101.3 J$ , the net heat absorbed by the gas in the two steps is

Answer: B

Initial

P V = 2 \times 10 = 20 L atm

and final

P V = 1 \times 20 = 20 L atm

, so the final temperature equals the initial one and

Δ U = 0

(ideal gas). Work is done only in step 1:

w = - P Δ V = - 2 \times 10 = - 20 L atm = - 2026 J

. Hence

q = - w = + 2026 J