One mole of an ideal diatomic gas $(C_{v} = 5 cal K^{- 1} {mol}^{- 1})$ changes its state from state 1 (27° C 1 L) to state 2 (127° C, 10 L). For this process. which of the following is(are) correct?
(Given : $R = 2 cal K^{- 1} {mol}^{- 1}$ )

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$ΔU = 500 cal$

$Δ S = 5$ in $(\frac{4}{3}) + 2$ in $10 cal K^{- 1}$

$ΔH = 700 cal$

$∆ G$ of the process can't be calculated using given information.

answer is A, B, C, D.

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Detailed Solution

$\begin{array}{l} ΔH = {nC}_{P} ΔT \\ = 1 \times (5 + 2) \times 100 = 700 cal \\ ΔU = {nC}_{V} ΔT \\ = 1 \times 5 \times 100 = 500 cal \\ ΔS = n_{R} \ln \frac{V_{2}}{V_{1}} + {nC}_{V} \ln \frac{T_{2}}{T_{1}} \\ = 2 \ln 10 + 5 \ln \frac{4}{3} \end{array}$