1 mole of an ideal gas $A (C_{v, m} = 3 R)$ and 2 mole of an ideal B are $(C_{v, m} = \frac{3}{2} R)$ , taken in a container and expanded reversible and adiabatically from 1 litre to 4 litre starting from initial temperature of 320K. $ΔE or ΔU$ for the process is

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-240 R

-960 R

-1920 Cal

240 R

answer is C.

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Detailed Solution

$\begin{array}{r} {Cv}_{(avg)} = \frac{1 \times 3 R \times 2 \times \frac{3}{2} R}{1 + 2} = 2 R \Rightarrow γ = 3 / 2 \\ T_{1} V_{1}^{γ - 1} = T_{2} V_{2}^{γ - 1} \\ 320 \times (1)^{1 / 2} = T_{2} (4)^{1 / 2} \Rightarrow T_{2} = \frac{320}{2} = 160 K \\ ΔU = {nC}_{v} ΔT = 3 \times 2 R \times (160 - 320) \\ = - 960 R = - 1920 cal \end{array}$