Question

a gas expands and does 450 j of work on the surroundings. at the same time, it absorbs 300 j of heat. a student attempts to calculate the change in internal energy δu as follows: check carefully the student’s work: 1. work done by the system ⇒ w(sys)= +450 2. heat absorbed ⇒ q=+300 j 3. δu = w+q determine which one of the following statements is correct? the sign of the work in step 1 is wrong the sign of the heat in step 2 is wrong the equation in step 3 is wrong all the steps are correct

Explanation:

Step1: Define work sign convention

When a system does work on the surroundings, work is negative for the system. Here, the gas does 450 J of work on the surroundings, so $w=- 450\ J$, not $w = + 450\ J$ as in step 1 of the student's work.

Step2: Define heat sign convention

When a system absorbs heat, $q$ is positive. The gas absorbs 300 J of heat, so $q = + 300\ J$ (correct in student's step 2).

Step3: Recall internal - energy formula

The first - law of thermodynamics is $\Delta U=q + w$. The formula used in step 3 of the student's work is correct.

Answer:

The sign of the work in step 1 is wrong