Question

Heat of $20Kcal$ is supplied to the system and $8400J$ of external work is done on the system so that its volume decreases at constant pressure. The change in internal energy is ($J=4200J/kcal$)

• A. $9.24\times {10}^{4}J$
• B. $7.56\times {10}^{4}J$
• C. $8.4\times {10}^{4}J$
• D. $10.5\times {10}^{4}J$

Answer 1

The correct option is A $9.24\times {10}^{4}J$

$\Delta Q=20\times {10}^{3}cal$

$=20\times 4200J$

$=84\times {10}^{3}J$

$\Delta W=-8400J$

$\therefore \Delta V=\Delta Q-\Delta W$

$=84000+8400$

$=92400$

$=9.24\times {10}^4$

Hence,

option $\left(A\right)$ is correct answer.