Question

A certain volume of dry gas at NTP is expanded reversibly to three times its volume adibatically. Calculate the final temperature and pressure assuming ideal behaviour.$\frac{C_p}{C_v}$ for air = 1.4.

Given: $3^{0.4}=1.55,3^{1.4}=4.66$

• A. 176 K, 0.215 atm
• B. 154 K, 1.52 atm
• C. 254 K, 0.568 atm
• D. 454 K, 0.684 atm

Answer 1

The correct option is A 176 K, 0.215 atm

$T_1$ (at NTP) =273 K, $V_2=3V_1$ $P_1$ = 1 atm
For adiabatic expansion, we have $T{V}^{\gamma -1}$ = constant
So, $\frac{T_1}{T_2}=(\frac{V_2}{V_1}{)}^{\gamma -1}$
$\therefore$
$\frac{273}{T_2}=(\frac{3V_1}{V_1}{)}^{1.4-1}$
Calculating for $T_2$, we get $T_2$ = 176 K
Similarly, final pressure under adiabatic conditions
$\frac{P_1}{P_2}=(\frac{V_2}{V_1}{)}^{\gamma }$
So, $\frac{1}{P_2}=(\frac{3V_1}{V_1}{)}^{1.4}$
$P_2$ = 0.215 atm