First-Law Bookkeeping Around a Rectangular P–V Path from A to D: 21-Mat-A1 Thermodynamics, December 2018, Q1

Question 1: First-Law Bookkeeping Around a Rectangular P–V Path from A to D (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 ($C_v=\tfrac32R=12.47\ \text{J}\,\text{K}^{-1}\text{mol}^{-1}$, $C_p=\tfrac52R=20.79\ \text{J}\,\text{K}^{-1}\text{mol}^{-1}$, $R=8.314\ \text{J}\,\text{K}^{-1}\text{mol}^{-1}$).

Corner states (P in kPa, V in L)
State	P	V
A	100	10
B	100	30
C	50	10
D	50	30

Find. $q$, $w$ (work done ON the system), $\Delta E$ and $\Delta H$ for Steps 1–4 and for the two composite paths A→B→D (Path 1) and A→C→D (Path 2).

Figure 1 — The two paths from A to D on the P–V plane: Path 1 (blue) runs A→B (isobaric) then B→D (isochoric); Path 2 (orange) runs A→C (isochoric) then C→D (isobaric). Both paths share the same start and end states.

Approach. Since 1 kPa·L $=$ 1 J, every corner's $PV$ product is already in joules; on an isobaric leg $w=-P\Delta V$ and $q=\Delta H$, on an isochoric leg $w=0$ and $q=\Delta E$, and on every leg $\Delta E=\tfrac32(P_fV_f-P_iV_i)$, $\Delta H=\tfrac52(P_fV_f-P_iV_i)$ because $\Delta E=nC_v\Delta T=\tfrac32 nR\Delta T=\tfrac32\Delta(PV)$.

(a) Step 1 (A→B, isobaric, $P=100$ kPa). $\Delta(PV)=P_BV_B-P_AV_A=3000-1000=2000$ J, so $\Delta E=\tfrac32(2000)=\boxed{+3000\ \text{J}}$ and $\Delta H=\tfrac52(2000)=\boxed{+5000\ \text{J}}$. Work done on the system as it expands at constant $P$: $$w=-P_A(V_B-V_A)=-(100)(30-10)=\boxed{-2000\ \text{J}}.$$ From the first law, $q=\Delta E-w=3000-(-2000)=\boxed{+5000\ \text{J}}$ (matches $\Delta H$, as required at constant pressure).
(b) Step 2 (B→D, isochoric, $V=30$ L). $\Delta(PV)=P_DV_D-P_BV_B=1500-3000=-1500$ J, so $\Delta E=\tfrac32(-1500)=\boxed{-2250\ \text{J}}$ and $\Delta H=\tfrac52(-1500)=\boxed{-3750\ \text{J}}$. No volume change means $w=0$, so $$q=\Delta E-w=\boxed{-2250\ \text{J}}.$$
(c) Step 3 (A→C, isochoric, $V=10$ L). $\Delta(PV)=P_CV_C-P_AV_A=500-1000=-500$ J, so $\Delta E=\tfrac32(-500)=\boxed{-750\ \text{J}}$ and $\Delta H=\tfrac52(-500)=\boxed{-1250\ \text{J}}$. Again $w=0$, so $$q=\Delta E-w=\boxed{-750\ \text{J}}.$$
(d) Step 4 (C→D, isobaric, $P=50$ kPa). $\Delta(PV)=P_DV_D-P_CV_C=1500-500=1000$ J, so $\Delta E=\tfrac32(1000)=\boxed{+1500\ \text{J}}$ and $\Delta H=\tfrac52(1000)=\boxed{+2500\ \text{J}}$. Work done on the system: $$w=-P_C(V_D-V_C)=-(50)(30-10)=\boxed{-1000\ \text{J}}.$$ $$q=\Delta E-w=1500-(-1000)=\boxed{+2500\ \text{J}}.$$
(e) Path 1 (Step 1 $+$ Step 2). Summing the two legs: $$q_{1}=5000-2250=\boxed{+2750\ \text{J}},\qquad w_{1}=-2000+0=\boxed{-2000\ \text{J}},$$ $$\Delta E_{1}=3000-2250=\boxed{+750\ \text{J}},\qquad \Delta H_{1}=5000-3750=\boxed{+1250\ \text{J}}.$$
(f) Path 2 (Step 3 $+$ Step 4). Summing the two legs: $$q_{2}=-750+2500=\boxed{+1750\ \text{J}},\qquad w_{2}=0-1000=\boxed{-1000\ \text{J}},$$ $$\Delta E_{2}=-750+1500=\boxed{+750\ \text{J}},\qquad \Delta H_{2}=-1250+2500=\boxed{+1250\ \text{J}}.$$ Both paths give the SAME $\Delta E$ and $\Delta H$ (state functions, path-independent) but DIFFERENT $q$ and $w$ (path-dependent), exactly as the first law requires. ✓

Quantity	Value
Step 1 (A→B): $q,w,\Delta E,\Delta H$	+5000, −2000, +3000, +5000 J
Step 2 (B→D): $q,w,\Delta E,\Delta H$	−2250, 0, −2250, −3750 J
Step 3 (A→C): $q,w,\Delta E,\Delta H$	−750, 0, −750, −1250 J
Step 4 (C→D): $q,w,\Delta E,\Delta H$	+2500, −1000, +1500, +2500 J
Path 1 (A→B→D): $q,w,\Delta E,\Delta H$	+2750, −2000, +750, +1250 J
Path 2 (A→C→D): $q,w,\Delta E,\Delta H$	+1750, −1000, +750, +1250 J

Concept & Theory

Topic. First law of thermodynamics applied leg-by-leg to a two-path ideal-gas process on a $P$–$V$ diagram.

Key relations. $PV=nRT$; $\Delta E=nC_v\Delta T=\tfrac32\Delta(PV)$; $\Delta H=nC_p\Delta T=\tfrac52\Delta(PV)$; $w_{on}=-P\Delta V$ (isobaric), $w_{on}=0$ (isochoric); $\Delta E=q+w_{on}$.

Why this works. For an ideal gas $E$ and $H$ are state functions of $T$ alone, so writing $\Delta(PV)=nR\Delta T$ turns every corner-to-corner change into an exact multiple of $\Delta(PV)$ — no path information is needed for $\Delta E$ or $\Delta H$. $q$ and $w$, by contrast, are path functions: the two routes from A to D split the same total energy change into different heat/work partitions.

Common pitfall. Assuming $q$ or $w$ must be equal for the two paths because $\Delta E$ and $\Delta H$ are — only the state functions are path-independent; treating $q$ or $w$ as state functions gives the wrong Step totals whenever a path is not a single isothermal or isobaric leg.

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

Question 1 of 7: First-Law Bookkeeping Around a Rectangular P–V Path from A to D

Question 1: First-Law Bookkeeping Around a Rectangular P–V Path from A to D (20 marks)