Question

A certain volume of dry air at NTP is expanded reversibly to four times its volume (a) isothermally (b) adiabatically. Calculate the final pressure and temperature in each case, assuming ideal behaviour.
$(\frac{C_P}{C_V}forair=1.4)$

Answer 1

Let $V_1$ be the initial volume of dry air at NTP
(a) Isothermally expansion: During isothermal expansion, the temperature remains the same throughout. Hence, final temperature will be $273K$.
Since, $P_1{V}_1=P_2{V}_2$
$P_2=\frac{P_1{V}_1}{V_2}=\frac{1\times V_1}{4V_1}=0.25atm$
(b) Adiabatic expansion:
$\frac{T_1}{T_2}={\left(\frac{V_2}{V_1}\right)}^{\gamma -1}$
$\frac{273}{T_2}={\left(\frac{4V_1}{V_1}\right)}^{1.4-1}=4^{0.4}$
$T_2=\frac{273}{4^{0.4}}=156.79K$
Final Pressure:
$\frac{P_1}{P_2}={\left(\frac{V_2}{V_1}\right)}^{\gamma }\frac{1}{P_2}={\left(\frac{4V_1}{V_1}\right)}^{1.4}=4^{1.4}$
$P_2=\frac{1}{4^{1.4}}=0.143atm$