The surroundings do 120 J of work on a gas. At the same time, the gas releases 23 J of heat. What is ΔE for this process?

02A

Question

The surroundings do 120 J of work on a gas. At the same time, the gas releases 23 J of heat. What is ΔE for this process?

02A

BlackTom AI · Accepted Answer

To solve, we use the first law of thermodynamics: ΔE = q + w, where q is heat added to the system and w is work done on the system (positive when done on the system).
First, assign signs from the problem: the surroundings do 120 J of work on the gas, so the work on the system w = +120 J. The gas releases 23 J of heat, which means heat leaves the system, so q = -23 J.
Now compute ΔE: ΔE = q + w = (-23 J) + (+120 J) = +97 J.
Option-by-option analysis:
-1) –143 J: This would imply a large decrease in internal energy, which would require either a large negative q or w; given q is negative but only -23 J and w is +120 J, this value is inconsistent with the energy balance. It also mixes signs in a way that doesn’t align with the first law.
-2) 143 J: This overestimates the energy change by effectively treating heat release as if it added energy or undercounting the work done on the system; the correct combination of q and w does not sum to +143 J.
-3) –97 J: This is the negative of the actual result. It would imply the system lost energy overall, which would require the net effect of q and w to be negative, not matching q = -23 J and w = +120 J.
-4) 97 J: This matches the calculation ΔE = (-23) + 120 = 97 J. The positive sign indicates the internal energy increased.
In summary, evaluating the signs carefully and applying ΔE = q + w yields ΔE = +97 J, which corresponds to option 4.

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The surroundings do 120 J of work on a gas. At the same time, the gas releases 23 J of heat. What is ΔE for this process? 02A

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