Question

Heat energy of 2100 J is given to a gas at a constant pressure of ${10}^5$ Pa. The increase in internal energy, if the change in volume is $5\times {10}^{-3}{m}^3$, is:

• A. 1500 J
• B. 1400 J
• C. 1600 J
• D. 800 J

Answer 1

Answer: (C)

1600 J

The work done to expand the gas is, $W$ $=P\Delta V$.
W $={10}^5\times 5\times {10}^{-3}$
Using the first law of thermodynamics,
$U=Q-W$
$U=2100-500$
$U=1600$.