Physics

Problem

One mole of an ideal gas at

300\,\text{K}

is taken through a reversible isothermal process. (a) During the first attempt, the gas expands and does

500\,\text{J}

of work. (b) In a second attempt at the same temperature, only

350\,\text{J}

of heat can be supplied from the reservoir; the desired isothermal condition is still to be maintained. Answer the following: a. (i) For part (a), state the value of

\Delta U

and explain your reasoning. b. (ii) For part (a), find the magnitude and algebraic sign of the heat exchanged

Q

. c. (iii) For part (b), determine the magnitude and algebraic sign of the mechanical work that must be done on the gas so that its temperature remains unchanged. d. (iv) Explain briefly how the signs chosen in parts (ii) and (iii) are consistent with the First Law.

Intuitive explanation

In an isothermal process, temperature stays constant, so internal energy doesn't change—meaning all the heat absorbed goes directly into work done by the gas. The key insight is that for an ideal gas at fixed $T$ , the work and heat are linked through $Q = W$ (the first law with $\Delta U = 0$ ). In part (a), you can find how much heat was absorbed; in part (b), the limited heat supply constrains how much work can be done, and you'll need to figure out what changes about the process itself—like the volume ratio or final pressure—while keeping $T$ constant.

Intuitive explanation

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