Work and internal-energy change of a gas with a P–V equation of state: 22-Mec-A1 Applied Thermodynamics and Heat Transfer, Undated paper, Q1

Question 1 — Work and internal-energy change of a gas with a P–V equation of state (Part A, equal value)

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.

Quantity	Symbol	Value
Equation of state	$P(v)$	$300v-200v^{2}$ kPa
Internal-energy relation	$u$	$0.3\,Pv+20$ kJ/kg
Initial specific volume	$v_1$	0.1 m³/kg
Final specific volume	$v_2$	0.4 m³/kg

Find. (a) boundary work $w$; (b) internal-energy change $\Delta u$; (c) heat transfer $q$ (per kg).

Boundary work is the area under the reconstructed $P$–$v$ curve between $v_1$ and $v_2$.

Approach. Integrate the given $P(v)$ for the closed-system boundary work, differentiate the $u$-relation across the end states for $\Delta u$, then close with the first law $q=\Delta u+w$.

Boundary work by integration. For a quasi-static process $w=\int_{v_1}^{v_2}P\,dv$ with $P=300v-200v^2$: $$w=\Big[150\,v^{2}-\tfrac{200}{3}v^{3}\Big]_{0.1}^{0.4}=\big(150(0.16)-66.7(0.064)\big)-\big(150(0.01)-66.7(0.001)\big)$$ $$w=(24.0-4.27)-(1.50-0.067)=\boxed{18.30\ \text{kJ/kg}}$$ The gas expands, so the work is done by the gas (positive).
Internal-energy change from the state relation. With $u=0.3Pv+20$, only the $Pv$ term changes, so $\Delta u=0.3\,(P_2v_2-P_1v_1)$. Evaluate the endpoints from $P(v)$: $P_1=300(0.1)-200(0.1)^2=28$ kPa and $P_2=300(0.4)-200(0.4)^2=88$ kPa. Hence $$\Delta u=0.3\big[(88)(0.4)-(28)(0.1)\big]=0.3\,(35.2-2.8)=\boxed{9.72\ \text{kJ/kg}}$$
Heat transfer (first law, closed system). $$q=\Delta u+w=9.72+18.30=\boxed{28.02\ \text{kJ/kg}}$$ Both energy stores increase, so heat is added to the gas.

Quantity	Result
Boundary work $w$	18.30 kJ/kg (by the gas)
Internal-energy change $\Delta u$	+9.72 kJ/kg
Heat transfer $q$	+28.02 kJ/kg (added)

Check — reconstructed coefficients.

The state-equation coefficients are given inconsistently ("0.3", "0.317", "0.2") and the question can be read either as a curve process or as a "constant 100 kPa" process. The self-consistent reconstruction used here, $P=300v-200v^2$ (kPa), reproduces exam-scale numbers ($w\sim18$ kJ/kg, $\Delta u\sim10$ kJ/kg); the method — integrate $P(v)$ for work, differentiate $u(Pv)$ for $\Delta u$, close with the first law — is what the question tests and is unaffected by the exact coefficients.

Concept & Theory — Q1

Topic: Closed-system boundary work for a non-linear $P$–$v$ path and the first law of thermodynamics.

Key relations: quasi-static work $w=\int P\,dv$ (Çengel, Ch. 4); first law for a closed system $q=\Delta u+w$; state postulate — any two independent intensive properties fix the state, here supplied as $P(v)$ and $u(P,v)$.

Why this works: The boundary work depends only on the path $P(v)$, which is given explicitly, so a direct integral suffices — no ideal-gas assumption is needed. The internal energy is a point function, so $\Delta u$ depends only on the end states; substituting the endpoint $Pv$ products into the given relation avoids integrating along the path.

Common pitfall: Treating $\Delta u=\int P\,dv$ or forgetting that $q$ and $w$ are path functions while $u$ is a state function — the first law is what links them.

Question 1 of 8: Work and internal-energy change of a gas with a P–V equation of state

Question 1 — Work and internal-energy change of a gas with a P–V equation of state (Part A, equal value)