First-Law Bookkeeping for a Monatomic Ideal Gas — Isobaric–Isochoric–Isothermal Cycle: 21-Mat-A1 Thermodynamics, May 2015, Q1

Question 1: First-Law Bookkeeping for a Monatomic Ideal Gas — Isobaric–Isochoric–Isothermal Cycle (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $n=1$ mol monatomic ideal gas, $C_v=20.8\ \text{J}\,\text{K}^{-1}\text{mol}^{-1}$ (so $C_p=C_v+R=29.11\ \text{J}\,\text{K}^{-1}\text{mol}^{-1}$). State A: $P_A=100$ kPa, $V_A=25$ L. Step 1 (isobaric) to state B: $P_B=100$ kPa, $V_B=100$ L. Step 2 (isochoric) to state C: $V_C=100$ L. Step 3 (isothermal) returns C→A, which fixes $T_C=T_A$.

Find. $q$, $w$ (on the system), $\Delta E\,(=\Delta U)$ and $\Delta H$ for each of the three steps and for the closed cycle.

Figure 1 — The A→B→C→A cycle in the P–V plane, drawn to the state values found in Step 1 below (isobar, isochor, then the $P=2500/V$ isotherm closing back to A).

Approach. Pin down every state's temperature and pressure from the ideal-gas law (using $T_C=T_A$ from Step 3's own description), then apply the standard constant-P, constant-V and isothermal-reversible work/heat relations leg by leg, and sum for the cycle.

State temperatures and $P_C$. $T_A=\dfrac{P_AV_A}{nR}=\dfrac{(100{,}000)(0.025)}{8.314}=300.7$ K; $T_B=\dfrac{P_BV_B}{nR}=\dfrac{(100{,}000)(0.100)}{8.314}=1202.8$ K. Step 3 returns isothermally to A, so $T_C=T_A=300.7$ K, and $$P_C=\frac{nRT_C}{V_C}=\frac{(8.314)(300.7)}{0.100}=\boxed{25.0\ \text{kPa}}.$$
(a) Step 1 (A→B, isobaric). $\Delta T=T_B-T_A=902.1$ K. $$w=-P_A\Delta V=-(100\ \text{kPa})(75\ \text{L})=-7500\ \text{J},\qquad q=nC_p\Delta T=(29.11)(902.1)=26{,}263.5\ \text{J},$$ $$\Delta U=nC_v\Delta T=(20.8)(902.1)=18{,}763.5\ \text{J},\qquad \Delta H=q=\boxed{26{,}263.5\ \text{J}}\ \ (\text{constant }P).$$ Check: $q+w=26{,}263.5-7500=18{,}763.5\ \text{J}=\Delta U$. ✓
(b) Step 2 (B→C, isochoric). $\Delta T=T_C-T_B=300.7-1202.8=-902.1$ K, and $w=0$ (no volume change). $$q=\Delta U=nC_v\Delta T=(20.8)(-902.1)=\boxed{-18{,}763.5\ \text{J}},\qquad \Delta H=nC_p\Delta T=(29.11)(-902.1)=-26{,}263.5\ \text{J}.$$
(c) Step 3 (C→A, isothermal, reversible). Ideal gas at constant $T$: $\Delta U=\Delta H=0$. Work done on the system during the reversible compression $V_C\to V_A$: $$w=-nRT_C\ln\!\frac{V_A}{V_C}=-(8.314)(300.7)\ln\!\frac{25}{100}=-2500\ln(0.25)=\boxed{+3465.7\ \text{J}}.$$ Since $\Delta U=0$, $q=-w=\boxed{-3465.7\ \text{J}}$ (heat is rejected as the gas is compressed isothermally).
(d) The cycle. Summing the three legs, and separately confirming $\Delta U$ and $\Delta H$ close to zero because A→B→C→A returns to the identical starting state: $$q_{cyc}=26{,}263.5-18{,}763.5-3465.7=\boxed{+4034.3\ \text{J}},\qquad w_{cyc}=-7500+0+3465.7=\boxed{-4034.3\ \text{J}},$$ $$\Delta U_{cyc}=\Delta H_{cyc}=\boxed{0}.$$ Check: $q_{cyc}+w_{cyc}=4034.3-4034.3=0=\Delta U_{cyc}$. ✓

Step	q (J)	w (J, on system)	ΔU (J)	ΔH (J)
1 (A→B, isobaric)	+26,263.5	−7,500.0	+18,763.5	+26,263.5
2 (B→C, isochoric)	−18,763.5	0	−18,763.5	−26,263.5
3 (C→A, isothermal)	−3,465.7	+3,465.7	0	0
Cycle	+4,034.3	−4,034.3	0	0

Concept & Theory

Topic. First law of thermodynamics applied leg-by-leg to a three-step ideal-gas cycle (isobaric–isochoric–isothermal).

Key relations. $\Delta U=nC_v\Delta T$; $\Delta H=nC_p\Delta T$; $w_{on}=-P\Delta V$ (isobaric); $w_{on}=-nRT\ln(V_f/V_i)$ (isothermal, reversible); $\Delta U=q+w_{on}$; $C_p-C_v=R$.

Why this works. Because $U$ and $H$ of an ideal gas depend on temperature alone, any process linking the same two temperatures gives identical $\Delta U$, $\Delta H$ regardless of path — that symmetry is why Step 2's $\Delta U$ and $\Delta H$ are exactly the negatives of Step 1's once $T_C=T_A$ is recognised. The cycle closing to $\Delta U_{cyc}=\Delta H_{cyc}=0$ is not a coincidence but a direct consequence of A, B, C, A being the same physical state visited twice — a built-in arithmetic check.

Common pitfall. Missing that "constant-temperature process back to State A" is what fixes $T_C=T_A$ (state C's own $T$ is not given directly by the Step 2 description) — without that link neither $P_C$ nor any Step 2 quantity can be found. Also reversing the sign convention for $w$ (work done on the system, not by it).

Source. Gaskell, Introduction to the Thermodynamics of Materials, Ch. 3–4 (first law, ideal-gas state functions and reversible work).

Question 1 of 7: First-Law Bookkeeping for a Monatomic Ideal Gas — Isobaric–Isochoric–Isothermal Cycle

Question 1: First-Law Bookkeeping for a Monatomic Ideal Gas — Isobaric–Isochoric–Isothermal Cycle (20 marks)