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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