According to the energy balance ΔU = Q − W, what happens to the internal energy if the gas absorbs heat and does not do any work?

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Multiple Choice

According to the energy balance ΔU = Q − W, what happens to the internal energy if the gas absorbs heat and does not do any work?

Explanation:
The change in internal energy follows the first law: ΔU = Q − W, where Q is heat added to the system and W is work done by the system. If the gas absorbs heat and does no work, then W = 0. Substituting gives ΔU = Q − 0 = Q. So the internal energy increases by the amount of heat that flowed in (and, for an ideal gas, this corresponds to a rise in temperature). The other forms don’t match the energy balance: they either combine terms incorrectly or claim independence from Q and W, which isn’t true.

The change in internal energy follows the first law: ΔU = Q − W, where Q is heat added to the system and W is work done by the system. If the gas absorbs heat and does no work, then W = 0. Substituting gives ΔU = Q − 0 = Q. So the internal energy increases by the amount of heat that flowed in (and, for an ideal gas, this corresponds to a rise in temperature). The other forms don’t match the energy balance: they either combine terms incorrectly or claim independence from Q and W, which isn’t true.

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