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Internal Energy

The internal ...

Question

The internal energy (U) of an ideal gas is plotted against volume for a cyclic process ABCDA, as shown in the figure.
The temperature of the gas at B and C are $500 K$ and $300 K$ , respectively. The heat absorbed by the gas (in cal/mol) in this cyclic process, is :

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Solution

We know,
$q = q_{A B} + q_{B C} + q_{C D} + q_{D A}$
$= n R T_{B} l n \frac{2 V_{0}}{V_{0}} + n C_{v, m} (T_{c} - T_{B}) + n R T_{c} l n \frac{V_{0}}{2 V_{0}} + n C_{v}, m (T_{A} - T_{B})$
$= n R (T_{B} - T_{c}) l n 2 = 1 \times 2 \times 0.7 \times 200$
$= 280$ cal/mol

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