The Reversible Adiabatic (Isentropic) Ideal-Gas Process: 21-Mat-A1 Thermodynamics, May 2015, Q2

Question 2: The Reversible Adiabatic (Isentropic) Ideal-Gas Process (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) $n$ mol ideal gas, reversible adiabatic process, $C_v$ constant, $C_p-C_v=R$. (b) $P_1=80$ kPa, $P_2=60$ kPa, $T_1=300$ K, $C_p=28.9\ \text{J}\,\text{K}^{-1}\text{mol}^{-1}$.

Find. (a) Show $P_iV_i^{\gamma}=P_fV_f^{\gamma}$. (b) $T_2$ after the adiabatic expansion.

Approach. Combine the first law (adiabatic, $\delta q=0$) with the ideal-gas equation of state to get a separable relation between $T$ and $V$; integrate, then substitute $T=PV/nR$ to re-express purely in $P$ and $V$. Part (b) reuses the same derivation with $V$ eliminated instead, giving the direct $T$–$P$ form.

(a) First law, adiabatic. $\delta q=0\Rightarrow dU=\delta w_{on}=-PdV$ (reversible). With $dU=nC_vdT$ (ideal gas, $C_v$ constant) and $P=nRT/V$ (ideal-gas law): $$nC_v\,dT=-\frac{nRT}{V}dV\ \Longrightarrow\ C_v\frac{dT}{T}=-R\frac{dV}{V}.$$
Integrate between the initial and final states. $$C_v\ln\frac{T_f}{T_i}=-R\ln\frac{V_f}{V_i}\ \Longrightarrow\ \left(\frac{T_f}{T_i}\right)^{C_v}=\left(\frac{V_i}{V_f}\right)^{R}.$$
Eliminate $T$ using $T=PV/(nR)$. $\dfrac{T_f}{T_i}=\dfrac{P_fV_f}{P_iV_i}$, so $$\left(\frac{P_f}{P_i}\right)^{C_v}\left(\frac{V_f}{V_i}\right)^{C_v}=\left(\frac{V_i}{V_f}\right)^{R}\ \Longrightarrow\ \left(\frac{P_f}{P_i}\right)^{C_v}=\left(\frac{V_i}{V_f}\right)^{R+C_v}=\left(\frac{V_i}{V_f}\right)^{C_p}$$ using $R+C_v=C_p$. Raising both sides to the power $1/C_v$ and writing $\gamma=C_p/C_v$: $$\frac{P_f}{P_i}=\left(\frac{V_i}{V_f}\right)^{\gamma}\ \Longrightarrow\ \boxed{P_iV_i^{\gamma}=P_fV_f^{\gamma}}.\ \blacksquare$$
(b) The equivalent T–P form. Eliminating $V$ instead of $P$ from the same two starting relations gives, by the identical steps, $$\left(\frac{T_f}{T_i}\right)^{C_p}=\left(\frac{P_f}{P_i}\right)^{R}\ \Longrightarrow\ T_f=T_i\left(\frac{P_f}{P_i}\right)^{R/C_p}.$$ Here $C_v=C_p-R=28.9-8.314=20.586\ \text{J}\,\text{K}^{-1}\text{mol}^{-1}$, $\gamma=C_p/C_v=1.404$, and $R/C_p=8.314/28.9=0.2877$.
Substitute the numbers. $$T_2=300\left(\frac{60}{80}\right)^{0.2877}=300(0.75)^{0.2877}=300(0.9206)=\boxed{276.2\ \text{K}}.$$ The gas cools on adiabatic expansion, as expected ($T_2

Part	Result
(a)	$P_iV_i^{\gamma}=P_fV_f^{\gamma}$ (derived)
(b) Final temperature $T_2$	276.2 K

Concept & Theory

Topic. Reversible adiabatic (isentropic) processes in an ideal gas and the three equivalent $PV^{\gamma}$, $TV^{\gamma-1}$, $TP^{(1-\gamma)/\gamma}$ relations.

Key relations. $dU=nC_vdT$; $PV=nRT$; $C_p-C_v=R$; $\gamma=C_p/C_v$; $PV^{\gamma}=\text{const}$, $TV^{\gamma-1}=\text{const}$, $TP^{(1-\gamma)/\gamma}=\text{const}$.

Why this works. "Reversible and adiabatic" is exactly the statement that entropy is constant ($\delta q_{rev}=TdS=0$), so combining the first law with the equation of state is a direct route to the isentropic path — no separate entropy argument is needed. All three forms of the adiabatic relation are algebraically equivalent; which one is fastest to use just depends on which two variables the problem specifies.

Common pitfall. Using $C_v$ where the $T$–$P$ form needs $C_p$ in the exponent (and vice-versa for the $T$–$V$ form) — the $T$–$V$ exponent is $R/C_v=\gamma-1$, the $T$–$P$ exponent is $R/C_p=(\gamma-1)/\gamma$.

Source. Gaskell, Introduction to the Thermodynamics of Materials, Ch. 4 (reversible adiabatic processes).

Question 2 of 7: The Reversible Adiabatic (Isentropic) Ideal-Gas Process

Question 2: The Reversible Adiabatic (Isentropic) Ideal-Gas Process (20 marks)