Gujarati
11.Thermodynamics
medium

The pressure in the tyre of a car is four times the atmospheric pressure at $300 K$. If this tyre suddenly bursts, its new temperature will be $(\gamma = 1.4)$

A

$300\,{(4)^{1.4/0.4}}$

B

$300\,{\left( {\frac{1}{4}} \right)^{ - 0.4/1.4}}$

C

$300\,{(2)^{ - 0.4/1.4}}$

D

$300\,{(4)^{ - 0.4/1.4}}$

Solution

(d) For adiabatic process $\frac{{{T^\gamma }}}{{{P^{\gamma – 1}}}} = $ constant
==>$\frac{{{T_2}}}{{{T_1}}} = {\left( {\frac{{{P_1}}}{{{P_2}}}} \right)^{\frac{{1 – \gamma }}{\gamma }}}$

==>$\frac{{{T_2}}}{{300}} = {\left( {\frac{4}{1}} \right)^{\frac{{(1 – 1.4)}}{{1.4}}}}$

==>${T_2} = 300{(4)^{ – \frac{{0.4}}{{1.4}}}}$

Standard 11
Physics

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