First-Law Bookkeeping for a Monatomic Ideal Gas — Two Paths, Same Endpoints: 21-Mat-A1 Thermodynamics, December 2015, Q1

Question 1: First-Law Bookkeeping for a Monatomic Ideal Gas — Two Paths, Same Endpoints (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 ($C_v=\tfrac32R=12.47\ \text{J}\,\text{K}^{-1}\text{mol}^{-1}$, $C_p=\tfrac52R=20.79\ \text{J}\,\text{K}^{-1}\text{mol}^{-1}$). State A: $P_A=200$ kPa, $V_A=10$ L. Path 1 — Step 1 (isobaric) A→B: $P_B=200$ kPa, $V_B=40$ L; Step 2 (isochoric) B→D: $P_D=50$ kPa, $V_D=40$ L. Path 2 — Step 3 (isochoric) A→C: $P_C=50$ kPa, $V_C=10$ L; Step 4 (isobaric) C→D.

Find. $q$, $w$ (on the system), $\Delta E\,(=\Delta U)$ and $\Delta H$ for each of the four steps, and for each of the two full paths A→D.

Figure 1 — The two paths A→D in the P–V plane: Path 1 (solid blue) via B at constant P then constant V; Path 2 (dashed red) via C at constant V then constant P.

Approach. Find every state's temperature from $PV=nRT$, then apply the constant-$P$ and constant-$V$ work/heat relations leg by leg; sum the two legs of each path and compare — since A and D share the same $PV$ product, $T_D=T_A$, which forces $\Delta U$ and $\Delta H$ to vanish over either full path.

State temperatures. $T_A=\dfrac{P_AV_A}{nR}=\dfrac{(200{,}000)(0.010)}{8.314}=240.6$ K; $T_B=\dfrac{(200{,}000)(0.040)}{8.314}=962.2$ K; $T_C=\dfrac{(50{,}000)(0.010)}{8.314}=60.1$ K; $T_D=\dfrac{(50{,}000)(0.040)}{8.314}=240.6$ K. Note $P_AV_A=P_DV_D=2000$ J, so $$\boxed{T_D=T_A=240.6\ \text{K}}$$ — A and D lie on the same isotherm even though neither pressure nor volume matches.
(a) Step 1 (A→B, isobaric). $\Delta T=T_B-T_A=721.7$ K. $$w=-P_A\Delta V=-(200\ \text{kPa})(30\ \text{L})=\boxed{-6000\ \text{J}},\qquad q=nC_p\Delta T=(20.79)(721.7)=\boxed{15{,}000\ \text{J}},$$ $$\Delta U=nC_v\Delta T=(12.47)(721.7)=9000\ \text{J},\qquad \Delta H=q=15{,}000\ \text{J}\ \ (\text{constant }P).$$ Check: $q+w=15{,}000-6000=9000\ \text{J}=\Delta U$. ✓
(b) Step 2 (B→D, isochoric). $\Delta T=T_D-T_B=-721.7$ K, and $w=0$ (no volume change). $$q=\Delta U=nC_v\Delta T=(12.47)(-721.7)=\boxed{-9000\ \text{J}},\qquad \Delta H=nC_p\Delta T=(20.79)(-721.7)=\boxed{-15{,}000\ \text{J}}.$$
(c) Step 3 (A→C, isochoric). $\Delta T=T_C-T_A=-180.4$ K, and $w=0$. $$q=\Delta U=nC_v\Delta T=(12.47)(-180.4)=\boxed{-2250\ \text{J}},\qquad \Delta H=nC_p\Delta T=(20.79)(-180.4)=\boxed{-3750\ \text{J}}.$$
(d) Step 4 (C→D, isobaric). $\Delta T=T_D-T_C=180.4$ K. $$w=-P_C\Delta V=-(50\ \text{kPa})(30\ \text{L})=\boxed{-1500\ \text{J}},\qquad q=nC_p\Delta T=(20.79)(180.4)=\boxed{3750\ \text{J}},$$ $$\Delta U=nC_v\Delta T=(12.47)(180.4)=2250\ \text{J},\qquad \Delta H=q=3750\ \text{J}.$$
(e) Path 1 (Step 1 + Step 2). $$q_{P1}=15{,}000-9000=\boxed{6000\ \text{J}},\qquad w_{P1}=-6000+0=\boxed{-6000\ \text{J}},$$ $$\Delta U_{P1}=9000-9000=\boxed{0},\qquad \Delta H_{P1}=15{,}000-15{,}000=\boxed{0}.$$ Check: $q_{P1}+w_{P1}=6000-6000=0=\Delta U_{P1}$. ✓
(f) Path 2 (Step 3 + Step 4). $$q_{P2}=-2250+3750=\boxed{1500\ \text{J}},\qquad w_{P2}=0-1500=\boxed{-1500\ \text{J}},$$ $$\Delta U_{P2}=-2250+2250=\boxed{0},\qquad \Delta H_{P2}=-3750+3750=\boxed{0}.$$ Both paths give $\Delta U=\Delta H=0$ (A and D are the same-temperature state), but $q$ and $w$ — path functions — differ four-fold between the two routes.

Step / Path	q (J)	w (J, on system)	ΔU (J)	ΔH (J)
1 (A→B, isobaric)	+15,000	−6,000	+9,000	+15,000
2 (B→D, isochoric)	−9,000	0	−9,000	−15,000
3 (A→C, isochoric)	−2,250	0	−2,250	−3,750
4 (C→D, isobaric)	+3,750	−1,500	+2,250	+3,750
Path 1 (A→B→D)	+6,000	−6,000	0	0
Path 2 (A→C→D)	+1,500	−1,500	0	0

Concept & Theory

Topic. First law of thermodynamics applied leg-by-leg to two different two-step ideal-gas paths sharing the same start and end states.

Key relations. $\Delta U=nC_v\Delta T$; $\Delta H=nC_p\Delta T$; $w_{on}=-P\Delta V$ (isobaric); $w_{on}=0$ (isochoric); $\Delta U=q+w_{on}$.

Why this works. $U$ and $H$ of an ideal gas depend on temperature alone, so any two paths connecting the same initial and final temperatures give identical $\Delta U$ and $\Delta H$ — here $P_AV_A=P_DV_D$ makes $T_D=T_A$, forcing both path totals to zero regardless of the intermediate states B or C. Heat and work are not state functions: they depend on the intermediate step taken, which is why Path 1 moves four times as much energy as heat and work as Path 2 even though both start and end identically.

Common pitfall. Assuming $q=0$ or $w=0$ over the whole path because $\Delta U=0$ — $\Delta U=0$ only fixes $q=-w$, not that either is individually zero; both must still be summed leg by leg.

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

Question 1 of 7: First-Law Bookkeeping for a Monatomic Ideal Gas — Two Paths, Same Endpoints

Question 1: First-Law Bookkeeping for a Monatomic Ideal Gas — Two Paths, Same Endpoints (20 marks)