Reversible Adiabatic Expansion of an Ideal Gas: 21-Mat-A1 Thermodynamics, December 2017, Q3

Question 3: Reversible Adiabatic Expansion of an Ideal Gas (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. (a) Reversible, adiabatic ($dq=0$), ideal gas, $C_v=$ const, $C_p-C_v=R$. (b) $P_1=80$ kPa, $P_2=60$ kPa, $T_1=300$ K, $C_p=28.9$ J K⁻¹mol⁻¹.

Find. (a) A derivation of $P_iV_i^\gamma=P_fV_f^\gamma$. (b) $T_2$.

Approach. (a) Combine the first law with $dq=0$ and the ideal-gas law to get a separable ODE in $T$ and $V$, integrate, and convert to $P$-$V$ form. (b) Use the $T$-$P$ form of the same adiabatic relation with $\gamma=C_p/C_v$.

(a) First law for a reversible adiabatic process. With $dq=0$, $dU=dq+dw=-P\,dV$ (work done on the system). For an ideal gas with constant $C_v$, $dU=nC_v\,dT$, so $$nC_v\,dT=-P\,dV=-\dfrac{nRT}{V}\,dV\quad\Rightarrow\quad C_v\dfrac{dT}{T}=-R\dfrac{dV}{V}.$$
(a) Integrate. Integrating between states $i$ and $f$: $$C_v\ln\!\left(\dfrac{T_f}{T_i}\right)=-R\ln\!\left(\dfrac{V_f}{V_i}\right)\quad\Rightarrow\quad \left(\dfrac{T_f}{T_i}\right)^{C_v}=\left(\dfrac{V_i}{V_f}\right)^{R}\quad\Rightarrow\quad T_iV_i^{R/C_v}=T_fV_f^{R/C_v}.$$ Since $C_p-C_v=R$, $R/C_v=\gamma-1$ where $\gamma=C_p/C_v$, giving $T_iV_i^{\gamma-1}=T_fV_f^{\gamma-1}$.
(a) Convert to $P$-$V$ form. Substituting $T=PV/nR$ into $TV^{\gamma-1}=\text{const}$: $$\dfrac{P_iV_i}{nR}V_i^{\gamma-1}=\dfrac{P_fV_f}{nR}V_f^{\gamma-1}\quad\Rightarrow\quad \boxed{P_iV_i^{\gamma}=P_fV_f^{\gamma}}.\ \blacksquare$$
(b) Ratio of heat capacities. $C_v=C_p-R=28.9-8.314=\boxed{20.586\ \text{J}\,\text{K}^{-1}\text{mol}^{-1}}$, so $\gamma=C_p/C_v=28.9/20.586=\boxed{1.404}$.
(b) Adiabatic $T$-$P$ relation. From $T^{\gamma}P^{1-\gamma}=\text{const}$ (equivalently $T_2/T_1=(P_2/P_1)^{(\gamma-1)/\gamma}$): $$T_2=T_1\left(\dfrac{P_2}{P_1}\right)^{(\gamma-1)/\gamma}=300\left(\dfrac{60}{80}\right)^{0.2878}=300(0.9206)=\boxed{276.2\ \text{K}}\ (3.2\,{}^\circ\text{C}).$$

Quantity	Value
(a) Adiabatic relation	$P_iV_i^\gamma=P_fV_f^\gamma$ (proved)
(b) $C_v$, $\gamma$	20.586 J K⁻¹mol⁻¹, 1.404
(b) $T_2$	276.2 K (3.2°C)

Concept & Theory

Topic. Reversible adiabatic (isentropic) processes in an ideal gas.

Key relations. $dU=nC_v\,dT=-P\,dV$ under $dq=0$; $TV^{\gamma-1}=\text{const}$; $PV^\gamma=\text{const}$; $T^\gamma P^{1-\gamma}=\text{const}$; $\gamma=C_p/C_v$.

Why this works. A reversible adiabatic path has zero entropy change, so the process is fully described by the ideal-gas equation of state plus the constraint that no heat crosses the boundary; combining the first law with $PV=nRT$ and integrating (holding $C_v$ constant) yields the classical power-law relation among any two of $P$, $V$, $T$.

Common pitfall. Using $C_p$ in place of $C_v$ in the $T$-$V$ derivation, or forgetting that expansion (here $P$ falling from 80 to 60 kPa) must cool an adiabatic ideal gas — a computed $T_2>T_1$ signals a sign or exponent error.

Source. Gaskell, Introduction to the Thermodynamics of Materials, Ch. 3 (reversible adiabatic processes, heat capacity ratio).

Question 3 of 7: Reversible Adiabatic Expansion of an Ideal Gas

Question 3: Reversible Adiabatic Expansion of an Ideal Gas (20 marks)