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Heat of 30 kcal is supplied to a system and 4200 J of external work is done on the system so that its volume decreases at constant pressure. What is the change in its internal energy. (J = 4200 J/kcal)

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$2.302 \times 10^{5} J$

$1.302 \times 10^{5} J$

$3.302 \times 10^{5} J$

$4.302 \times 10^{5} J$

answer is A.

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Detailed Solution

$\begin{array}{l} + dQ = 30 kcal : - dW = 4200 J : du = ? \\ J = 4200 J / kcal \\ dQ = 30 \times 4200 = 126000 J \\ dQ = du + dW \Rightarrow dQ = du - dw \\ du = dQ + dw = 126000 + 4200 \\ = 130200 = 1 .302 \times 10^{5} J \end{array}$

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