Two moles of a monoatomic ideal gas at $300 K$ occupy $10.0 L$ . The gas expands reversibly and adiabatically to $40.0 L$ . Calculate the magnitude of work done by the gas in $J$ (nearest integer). Use $R = 8.314 J mo l^{- 1} K^{- 1}$ .

Numerical answer type, answer in J, tolerance 1.

Answer: 4513.13 J

γ = 5/3

, so

T_{2} = T_{1} (V_{1} / V_{2})^{2/3} = 300 \times (1/4)^{2/3} = 119.06 K

. Since

q = 0

∣ w ∣ = n C_{v} (T_{1} - T_{2}) = 2 \times 1.5 \times 8.314 \times (300 - 119.06) = 4513.1 J