One mole of a monatomic ideal gas at $400 K$ expands reversibly along the path $P V^{2} = constant$ to twice its volume. Taking $R = 8.314 J mo l^{- 1} K^{- 1}$ , find the heat $q$ absorbed by the gas in J (nearest integer; give the sign).

Numerical answer type, answer in J, tolerance 1.

Answer: -831 J

From

P V^{2} = k

and

P V = RT

T V = constant

, so

T_{2} = 400/2 = 200 K

. Work by gas

= \int P d V = R (T_{1} - T_{2}) / (2 - 1) = 1662.8 J

Δ U = 1.5 R (T_{2} - T_{1}) = - 2494.2 J

. Then

q = Δ U + w_{b y} = - 2494.2 + 1662.8 = - 831.4 J

, about -831 J (heat is released).