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JEE Advanced 2010Paper 2
Exam
JEE Advanced
Year
2010
Paper
Paper 2
Subject
Physics
Chapter
Thermodynamics

JEE Advanced 2010 · Paper 2 · Physics · Thermodynamics

A diatomic ideal gas is compressed adiabatically to (1)/(32) of its...

A diatomic ideal gas is compressed adiabatically to 132\frac{1}{32} of its initial volume. If the initial temperature of the gas is TiT_i TiT_i (in kelvin) and the final temperature is aTia T_i, the value of aa is

Correct answer

0

Step-by-step solution

In adiabatic process, TVγ−1= constant ∴TiVi0.4=TfVf0.4 (as γ=1.4 for  diatomic  gas )  or \quadTiVi0.4=(\alphaTi)(Vi32)0.4 or  α(32)0.4=4\begin{aligned}& T V^{\gamma-1}=\text{ constant }\\ & \therefore T_i V_i^{0.4}=T_f V_f^{0.4} \\ & \text{ (as } \gamma=1.4 \text{\text{ for }\text{ diatomic }\text{ gas }) }\\ & \text{ or } \quadT_i V_i^{0.4}=\left(\alphaT_i\right)\left(\frac{V_i}{32}\right)^{0.4}\\ & \text{\text{ or } } \quad \alpha(32)^{0.4}=4 \\ & \end{aligned}\therefore$ The correct answer is 4 .
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