A gas absorbs 6 J of heat and does 4 J of work; what is the change in internal energy ΔU?

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Multiple Choice

A gas absorbs 6 J of heat and does 4 J of work; what is the change in internal energy ΔU?

Explanation:
Energy stored inside a gas changes when heat is added and when the gas does work on its surroundings. The first law of thermodynamics gives ΔU = Q − W, where Q is heat added to the system and W is the work done by the system. Here, the gas absorbs 6 J of heat, so Q = +6 J. It also does 4 J of work on the surroundings, so W = +4 J. Plugging in: ΔU = 6 − 4 = +2 J. So the internal energy increases by 2 J. (Sign conventions can swap depending on how W is defined, but using the standard Q − W form gives the +2 J result.)

Energy stored inside a gas changes when heat is added and when the gas does work on its surroundings. The first law of thermodynamics gives ΔU = Q − W, where Q is heat added to the system and W is the work done by the system.

Here, the gas absorbs 6 J of heat, so Q = +6 J. It also does 4 J of work on the surroundings, so W = +4 J. Plugging in: ΔU = 6 − 4 = +2 J.

So the internal energy increases by 2 J. (Sign conventions can swap depending on how W is defined, but using the standard Q − W form gives the +2 J result.)

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