Question

When 20 J of work was done on a gas, 40 J of heat energy was released. If the initial internal energy of the gas was 70 J. Then determine the final internal energy of the system.

• A. 50 J
• B. 60 J
• C. 90 J
• D. 10 J

Answer 1

The correct option is A 50 J

The correct answer is option(A).

Change in internal energy is:

$\Delta U=U_{final}-U_{initial}...........\left(i\right)$

According to first law of thermodynamics,

$\Delta U=Q-W........\left(ii\right)$

Aaccording to the question,

$Q=-Q$ (energy released in the system)

and, $W=-W$, (work done on the system).

So, equation (ii) becomes,

$U_{final}-U_{initial}=-Q+W$

$U_{final}=70-40+20=50J$