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A certain mass of gas at $273 K$ is expanded to $81$ times its volume under adiabatic condition. If $\gamma = 1.25$ for the gas, then its final temperature is ..... $^oC$
A
$-235$
B
$-182$
C
$-91$
D
$0$
Solution
(b) For adiabatic process $T{V^{\gamma – 1}}$= constant
==> $\frac{{{T_2}}}{{{T_1}}} = {\left( {\frac{{{V_1}}}{{{V_2}}}} \right)^{\gamma – 1}}$==> ${T_2} = {\left( {\frac{{{V_1}}}{{{V_2}}}} \right)^{\gamma – 1}} \times {T_1}$
==> ${T_2} = {\left( {\frac{1}{{81}}} \right)^{1.25 – 1}} \times 273$$ = {\left( {\frac{1}{{81}}} \right)^{0.25}} \times 273$
$ = \frac{{273}}{3} = 91K = \,-182°C$
Standard 11
Physics
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