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Q.

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            

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a

– 182°C

b

– 91°C

c

– 0°C

d

– 235°C

answer is B.

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Detailed Solution

For adiabatic process T{V^{\gamma - 1}} = constant
\Rightarrow\frac{{{T_2}}}{{{T_1}}} = {\left( {\frac{{{V_1}}}{{{V_2}}}} \right)^{\gamma - 1}}\Rightarrow\ {T_2} = {\left( {\frac{{{V_1}}}{{{V_2}}}} \right)^{\gamma - 1}} \times {T_1}
\Rightarrow{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 \rightarrow–182°C

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A certain mass of gas at 273 K is expanded to 81 times its volume under adiabatic condition. If   for the gas, then its final temperature is