Ranking: heat absorbed in four transformations

Problem

A monatomic ideal gas undergoes four transformations starting from the same initial state $(p, V, T)$. Rank the heat $Q$ exchanged from the smallest (which may be negative) to the largest.

• (A) Isotherm with a $50\%$ increase in volume.
• (B) Adiabat with a $50\%$ increase in volume.
• (C) Isobar with a $50\%$ increase in volume.
• (D) Isochor with a $50\%$ increase in pressure.

Solution

Setup. Monatomic gas: $f = 3$, so $C_v = \tfrac{3}{2}R$ and $C_p = \tfrac{5}{2}R$. Let us analyse each case using the first law $Q = \Delta U + L$ with $\Delta U = n C_v \Delta T$.

(B) Adiabat. By definition there is no heat exchange: $Q_B = 0$

(A) Isotherm. $\Delta U = 0$ (the temperature does not change), so $Q = L$: $Q_A = nRT\ln\frac{V_f}{V_i} = nRT\ln(1{,}5) = 0{,}405\,nRT$

(D) Isochor, $+50\%$ pressure. At constant volume $p \propto T$, so a $50\%$ increase in pressure implies $\Delta T = 0{,}5\,T$. The work is zero ($L = 0$): $Q_D = n C_v \Delta T = n\,\tfrac{3}{2}R\,(0{,}5\,T) = 0{,}75\,nRT$

(C) Isobar, $+50\%$ volume. At constant pressure $V \propto T$, so a $50\%$ increase in volume implies $\Delta T = 0{,}5\,T$: $Q_C = n C_p \Delta T = n\,\tfrac{5}{2}R\,(0{,}5\,T) = 1{,}25\,nRT$

Ranking. Comparing the coefficients $0 < 0{,}405 < 0{,}75 < 1{,}25$: $\ev{B < A < D < C}$

Note: no $Q$ is negative because every transformation involves expansion or heating starting from the initial state (Simonyi 2012).

Links

Topics: Thermodynamics Concepts: Thermodynamic transformations Skills: Ranking reasoning Objects: Ideal gas