Work In A Polytropic Cycle

Question

One mole of an ideal monatomic gas is heated at constant volume so its absolute temperature doubles from 300 K to 600 K. Taking $R=8.31	ext{ J mol}^{-1}	ext{K}^{-1}$, how much heat must be supplied?

RankGuru · Accepted Answer

Because the volume is held constant, the process is isochoric, so the gas does no work and the First Law reduces to $Q = \Delta \cup = nC_V\Delta T$. For a monatomic ideal gas $C_V = 	frac{3}{2}R$, since it has three translational degrees of freedom each contributing $	frac{1}{2}R$. Substituting $n = 1$, $C_V = 	frac{3}{2}(8.31) = 12.47	ext{ J mol}^{-1}	ext{K}^{-1}$, and $\Delta T = 600 - 300 = 300	ext{ K}$ gives $Q = 1	imes12.47	imes300 \approx 3740	ext{ J}$. The value 6232 J wrongly uses $C_P = 	frac{5}{2}R$, appropriate for constant pressure, not constant volume. The value 2494 J uses $R$ alone without the factor $	frac{3}{2}$. The value 1247 J halves the correct $C_V$ contribution. Since no work is done, all supplied heat raises internal energy, so $Q$ equals $\Delta U$ exactly. Had the same temperature rise been carried out at constant pressure instead, more heat would be required because part of it would go into expansion work, illustrating why $C_P$ exceeds $C_V$ by $R$. As a magnitude check, doubling the temperature of one mole of monatomic gas should add a few kilojoules of internal energy, and 3740 J sits squarely in that expected range.

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