Physics

Physics

Problem

During an isothermal expansion of an ideal gas from volume V1V_1 to V2V_2, which of the following correctly states the algebraic signs of QQ, WW and ΔU\Delta U? a) Q>0,  W<0,  ΔU=0Q>0,\;W<0,\;\Delta U=0 b) Q>0,  W>0,  ΔU=0Q>0,\;W>0,\;\Delta U=0 c) Q>0,  W>0,  ΔU>0Q>0,\;W>0,\;\Delta U>0 d) Q=0,  W>0,  ΔU<0Q=0,\;W>0,\;\Delta U<0
Intuitive explanation

In an isothermal process, temperature—and therefore internal energy—stays constant, so ΔU=0\Delta U = 0 always. The key insight is that the gas does work on the surroundings during expansion, making WW positive by convention, and the first law (Q=ΔU+WQ = \Delta U + W) then tells you QQ must also be positive to replenish that energy. Think of it as: the gas needs heat input to maintain its temperature while pushing outward.

Intuitive explanation

In an isothermal process, temperature—and therefore internal energy—stays constant, so ΔU=0\Delta U = 0 always. The key insight is that the gas does work on the surroundings during expansion, making WW positive by convention, and the first law (Q=ΔU+WQ = \Delta U + W) then tells you QQ must also be positive to replenish that energy. Think of it as: the gas needs heat input to maintain its temperature while pushing outward.

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