One mole of a monatomic ideal gas at $400 K$ expands from volume $V$ to $2 V$ along the path $P V^{1.2} = constant$ . With $R = 8.314 J mo l^{- 1} K^{- 1}$ , the heat exchanged by the gas is

Answer: C

With

P V^{1.2}

constant,

T V^{0.2}

is constant so

T_{2} = 400 \times 2^{- 0.2}

and

T_{1} - T_{2} = 51.8 K

(the gas cools).

Δ U = - 1.5 R (T_{1} - T_{2}) = - 646 J

, and

w_{b y} = R (T_{1} - T_{2}) / (1.2 - 1) = 2152 J

. By the first law

q = Δ U + w_{b y} = + 1507 J

: although

T

falls, the gas absorbs heat because

1.2 < γ

. Answer

- 646

is only

Δ U

, and

2152

is only the work.