Physics

Physics

Problem

An ideal gas undergoes a reversible isothermal expansion. a. (a) Starting from the definition of mechanical work during a volume change, derive an expression for the work done when nn moles of an ideal gas expand reversibly and isothermally from an initial volume ViV_i to a final volume VfV_f. b. (b) Two moles of the gas at 300 K300\,\text{K} expand reversibly from 0.002 m30.002\,\text{m}^3 to 0.010 m30.010\,\text{m}^3 in a single continuous step. Calculate the work done in this process. c. (c) The same overall change in volume is now carried out in two reversible isothermal stages at 300 K300\,\text{K}: first from 0.002 m30.002\,\text{m}^3 to 0.004 m30.004\,\text{m}^3, and then from 0.004 m30.004\,\text{m}^3 to 0.010 m30.010\,\text{m}^3. Show that the sum of the works in the two stages equals the work found in part (b). What does this result indicate about the dependence of work on the path for a reversible isothermal process?
Intuitive explanation

In an isothermal process, temperature stays constant, so the internal energy of an ideal gas doesn't change—all the heat absorbed goes directly into doing work on the surroundings. The key insight is that pV=constantpV = \text{constant} along this path, which means pressure drops as volume increases, making the work integral ∫p dV\int p \, dV solvable using logarithms rather than simple formulas. Understanding this process reveals why isothermal expansion is maximally efficient at extracting work from a gas.

Intuitive explanation

In an isothermal process, temperature stays constant, so the internal energy of an ideal gas doesn't change—all the heat absorbed goes directly into doing work on the surroundings. The key insight is that pV=constantpV = \text{constant} along this path, which means pressure drops as volume increases, making the work integral ∫p dV\int p \, dV solvable using logarithms rather than simple formulas. Understanding this process reveals why isothermal expansion is maximally efficient at extracting work from a gas.

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