Two moles of a monatomic ideal gas at $450 K$ and $10.0 L$ expand reversibly and adiabatically to $40.0 L$ . With $R = 8.314 J mo l^{- 1} K^{- 1}$ , the magnitude of work done by the gas, to the nearest joule, is

Numerical answer type, answer in J, tolerance 1.

Answer: 6770 J

T_{2} = T_{1} (V_{1} / V_{2})^{2/3} = 450 (1/4)^{2/3} = 178.58 K

∣ w ∣ = n C_{V} (T_{1} - T_{2}) = 2 (1.5) (8.314) (271.42) = 6769.7 J