Question

A system generates $50J$ of electrical energy, and delivers $150J$ of pressure-volume work done against the surroundings while releasing $300J$ of heat energy. What is the change in the internal energy of the system?

• A. $-100$
• B. $-400$
• C. $-300$
• D. $-500$

Answer 1

The correct option is D $-500$

Given:
$dU$ $=$ $dq$ $+$ $dw$
where, $dw$ $=$ ${dw}_{pv}$ $+$ ${dw}_{non-pv}$

So, we have $dU$ $=$ $dq$ $+$ ${dw}_{pv}$ $+$ ${dw}_{non-pv}$

Note: As per IUPAC convention, work done on the system is $positive$.
${dw}_{non-pv}$ $=$ $-50$ $J$ as the system has generated this electrical energy which means this is the work done by the system.
${dw}_{pv}$ $=$ $-150$ $J$ as the system has done work against the surroundings.
$dq$ $=$ $-300$ $J$ as the system released this as heat energy.
Therefore, from the above equation:
$dU$ $=$ $(-300)$ $J$ $+$ $(-150)$ $J$ $+$ $(-50)$ $J$
$dU$ $=$ $-500$ $J$

So, the correct answer is ${}^{\prime }{D}^{\prime }$.