0.5 mole each of two ideal gases $A (C_{v, m} = \frac{5}{2} R)$ and $B (C_{v, m} = 3 R)$ are taken in a container and expanded reversibly and adiabatically, during this process, temperature of gaseous mixture decreased from 350 K and 250 K Find $Δ H$ . (in cal/mol) for the process:

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-375R

-100R

-137.5 R

None of these

answer is C.

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Detailed Solution

$\begin{matrix} C_{v} (mix) = \frac{\frac{5}{2} R + 3 R}{2} = \frac{11 R}{4} \\ C_{p} = \frac{15 R}{4} \end{matrix} \begin{aligned} Δ H = n C_{p} Δ T \\ = - 1 \times \frac{15 R}{4} \times 100 \\ = - 375 R \end{aligned}$

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