Question

In a thermodynamic process, a gas releases 20 J of heat and 10 J of work is done on the gas. If initial internal energy was 40 J, the final internal energy is:

• A. 30 J
• B. 20 J
• C. 60 J
• D. 40 J

Hint:

Remember sign conventions: $Q > 0$ for heat added, $W > 0$ for work done by gas.

Answer 1

The Correct Option is A

Step 1: List the given values.
A gas releases 20 J of heat, and 10 J of work is done on the gas. Its starting internal energy is 40 J. We find the final internal energy.

Step 2: Recall the first law.
The first law of thermodynamics says $\Delta U = Q - W$, where $Q$ is heat added to the gas and $W$ is work done by the gas. We must use the correct signs.

Step 3: Fix the sign of heat.
Heat is released by the gas, so heat is leaving. That makes $Q = -20$ J.

Step 4: Fix the sign of work.
Work is done on the gas, not by it. So work done by the gas is negative: $W = -10$ J.

Step 5: Find the change in internal energy.
$\Delta U = Q - W = (-20) - (-10) = -20 + 10 = -10$ J. The internal energy drops by 10 J.

Step 6: Find the final energy.
$U_f = U_i + \Delta U = 40 + (-10) = 30$ J. \[ \boxed{30 \ \text{J}} \]