Problem — Adiabatic compression of a diatomic gas

Problem

A diatomic gas ($\gamma = 7/5$) is initially at $P_A = 10^5\;\text{Pa}$, $V_A = 10^{-2}\;\text{m}^3$, $T_A = 300\;\text{K}$ and undergoes a reversible adiabatic compression down to $V_B = V_A/2$. Calculate $P_B$ and $T_B$.

Solution

Setup. In a reversible adiabatic process $PV^\gamma = \text{const}$ and $TV^{\gamma-1} = \text{const}$ hold, with $\gamma = 7/5 = 1{,}4$. Since $V_B = V_A/2$, the volume ratio is $V_A/V_B = 2$.

Final pressure. From $P_A V_A^\gamma = P_B V_B^\gamma$:

$P_B = P_A\left(\frac{V_A}{V_B}\right)^{\gamma} = 10^5 \cdot 2^{1{,}4}$

With $2^{1{,}4} = e^{1{,}4\ln 2} = e^{0{,}970} \approx 2{,}639$:

$P_B \approx 10^5 \cdot 2{,}639 = \ev{2{,}64\cdot 10^5\;\text{Pa}}$

Final temperature. From $T_A V_A^{\gamma-1} = T_B V_B^{\gamma-1}$:

$T_B = T_A\left(\frac{V_A}{V_B}\right)^{\gamma-1} = 300 \cdot 2^{0{,}4}$

With $2^{0{,}4} = e^{0{,}4\ln 2} = e^{0{,}277} \approx 1{,}320$:

$T_B \approx 300 \cdot 1{,}320 = \ev{396\;\text{K}}$

Consistency check. With $PV = nRT$ we have $T_B/T_A = (P_B V_B)/(P_A V_A) = 2{,}639 \cdot (1/2) = 1{,}320$, i.e. $T_B = 396\;\text{K}$ $\checkmark$. Compressing the gas without any heat exchange raises both the pressure and the temperature: the work done on the gas increases its internal energy.

Collegamenti

Argomenti: Termodinamica Concetti: Trasformazioni termodinamiche · Legge dei gas perfetti Competenze: Impostazione simbolica Oggetti: Gas ideale · Pistone e cilindro