First-Law Bookkeeping Around a Rectangular P–V Cycle: 21-Mat-A1 Thermodynamics, May 2017, Q1

Question 1: First-Law Bookkeeping Around a Rectangular P–V 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 ($C_v=\tfrac32R=12.47\ \text{J}\,\text{K}^{-1}\text{mol}^{-1}$, $C_p=\tfrac52R=20.79\ \text{J}\,\text{K}^{-1}\text{mol}^{-1}$). Step 1 (isobaric, $P=200$ kPa): A(20 L)→B(50 L). Step 2 (isochoric, $V=50$ L): B(200 kPa)→C(100 kPa). Step 3 (isobaric, $P=100$ kPa): C(50 L)→D(20 L). Step 4 (isochoric, $V=20$ L): D(100 kPa)→A(200 kPa).

Find. $q$, $w$ (on the system), $\Delta E\,(=\Delta U)$ and $\Delta H$ for each of the four steps, and for the complete cycle A→B→C→D→A.

Figure 1 — The rectangular cycle in the P–V plane: Step 1 (A→B) and Step 3 (C→D) are isobaric legs; Step 2 (B→C) and Step 4 (D→A) are isochoric legs.

Approach. Find every corner's temperature from $PV=nRT$, then apply $w=-P\Delta V$ (isobaric) or $w=0$ (isochoric) with $q=nC_p\Delta T$ or $q=nC_v\Delta T$ leg by leg, and sum the four legs for the full-cycle totals.

Corner temperatures. $T=\dfrac{PV}{nR}$ gives $$T_A=\dfrac{(200{,}000)(0.020)}{8.314}=481.1\ \text{K},\quad T_B=\dfrac{(200{,}000)(0.050)}{8.314}=1202.8\ \text{K},$$ $$T_C=\dfrac{(100{,}000)(0.050)}{8.314}=601.4\ \text{K},\quad T_D=\dfrac{(100{,}000)(0.020)}{8.314}=\boxed{240.6\ \text{K}}.$$
(a) Step 1 (A→B, isobaric, $P=200$ kPa). $\Delta T=T_B-T_A=721.7$ K. $$w=-P\Delta V=-(200\ \text{kPa})(30\ \text{L})=\boxed{-6000\ \text{J}},\qquad q=nC_p\Delta T=(20.79)(721.7)=\boxed{15{,}000\ \text{J}},$$ $$\Delta U=nC_v\Delta T=(12.47)(721.7)=9000\ \text{J},\qquad \Delta H=q=15{,}000\ \text{J}\ \ (\text{constant }P).$$ Check: $q+w=15{,}000-6000=9000\ \text{J}=\Delta U$. ✓
(b) Step 2 (B→C, isochoric, $V=50$ L). $\Delta T=T_C-T_B=-601.4$ K, and $w=0$ (no volume change). $$q=\Delta U=nC_v\Delta T=(12.47)(-601.4)=\boxed{-7500\ \text{J}},\qquad \Delta H=nC_p\Delta T=(20.79)(-601.4)=\boxed{-12{,}500\ \text{J}}.$$
(c) Step 3 (C→D, isobaric, $P=100$ kPa). $\Delta T=T_D-T_C=-360.7$ K. $$w=-P\Delta V=-(100\ \text{kPa})(-30\ \text{L})=\boxed{3000\ \text{J}},\qquad q=nC_p\Delta T=(20.79)(-360.7)=\boxed{-7500\ \text{J}},$$ $$\Delta U=nC_v\Delta T=(12.47)(-360.7)=-4500\ \text{J},\qquad \Delta H=q=-7500\ \text{J}.$$
(d) Step 4 (D→A, isochoric, $V=20$ L). $\Delta T=T_A-T_D=240.6$ K, and $w=0$. $$q=\Delta U=nC_v\Delta T=(12.47)(240.6)=\boxed{3000\ \text{J}},\qquad \Delta H=nC_p\Delta T=(20.79)(240.6)=\boxed{5000\ \text{J}}.$$
(e) Complete cycle (Steps 1–4 summed). $$q_{cyc}=15{,}000-7500-7500+3000=\boxed{3000\ \text{J}},\qquad w_{cyc}=-6000+0+3000+0=\boxed{-3000\ \text{J}},$$ $$\Delta U_{cyc}=9000-7500-4500+3000=\boxed{0},\qquad \Delta H_{cyc}=15{,}000-12{,}500-7500+5000=\boxed{0}.$$ Check: $q_{cyc}+w_{cyc}=3000-3000=0=\Delta U_{cyc}$. ✓ Both state functions vanish over any closed cycle back to A, while $q$ and $w$ — path functions — net to a non-zero value equal to the work done by the gas on its surroundings.

Step / Cycle	q (J)	w (J, on system)	ΔU (J)	ΔH (J)
1 (A→B, isobaric)	+15,000	−6,000	+9,000	+15,000
2 (B→C, isochoric)	−7,500	0	−7,500	−12,500
3 (C→D, isobaric)	−7,500	+3,000	−4,500	−7,500
4 (D→A, isochoric)	+3,000	0	+3,000	+5,000
Full cycle	+3,000	−3,000	0	0

Concept & Theory

Topic. First law of thermodynamics applied leg-by-leg to a rectangular ideal-gas P–V cycle.

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 are functions of temperature alone, so any closed cycle that returns to its starting state has $\Delta U_{cyc}=\Delta H_{cyc}=0$ regardless of the path taken between corners. Heat and work are not state functions — they depend on the sequence of legs — so the cycle nets a non-zero $q_{cyc}=-w_{cyc}=3000$ J, equal to the area enclosed by the rectangle in the P–V plane (the net work the gas cycle performs).

Common pitfall. Forgetting that $w=0$ only on the isochoric legs, not throughout — students sometimes apply $w=-P\Delta V$ to every step and get a spurious non-zero $w$ for Steps 2 and 4.

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

Question 1 of 7: First-Law Bookkeeping Around a Rectangular P–V Cycle

Question 1: First-Law Bookkeeping Around a Rectangular P–V Cycle (20 marks)