Two moles of a monatomic ideal gas ( $C_{V} = \frac{3}{2} R$ ) change reversibly from $300 K$ and $10.0 L$ to $450 K$ and $40.0 L$ . Using $R = 8.314 J mo l^{- 1} K^{- 1}$ , find $Δ S$ of the gas in $J K^{- 1}$ (two decimals).

Numerical answer type, answer in J/K, tolerance 0.05.

Answer: 33.16 J/K

Δ S = n C_{V} ln (T_{2} / T_{1}) + n R ln (V_{2} / V_{1})

. The thermal term is

2 \times 1.5 \times 8.314 \times ln 1.5 = 10.11

and the volume term is

2 \times 8.314 \times ln 4 = 23.05

. Sum

= 33.16 J K^{- 1}