Question

Air is filled in a tube of the wheel of a car at ${27}^{\circ }C$ and 2 atm pressure if the tube is suddenly bursts, the final temperature of air will be-
$(\gamma =1.5,2^{1/3}=1.251)$

• A. $-{33}^{\circ }C$
• B. $0^{\circ }C$
• C. ${21.6}^{\circ }C$
• D. ${240}^{\circ }C$

Answer 1

The correct option is D ${240}^{\circ }C$

Given,

$T_1=27+273=300K$

$P_1=2atm$

$\gamma =1.5$ $2^{1/3}=1.251$

In the process

$T_1^{\gamma }{P}_1^{\gamma -1}=T_2^{\gamma }{P}_2^{\gamma -1}$

$(\frac{T_1}{T_2}{)}^{\gamma }=(\frac{P_2}{P_1}{)}^{\gamma -1}$

$(\frac{300}{T_2}{)}^{3/2}=(\frac{2}{1}{)}^{1/2}$

$T_2=\frac{300}{2^{1/3}}={240}^{0}C$

The correct option is D.