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JIPMER 2006
Exam
JIPMER
Year
2006
Subject
Physics
Chapter
Thermodynamics

JIPMER 2006 · Physics · Thermodynamics

A perfect gas is found to obey the relation P V^ 3 / 2 = constant...

A perfect gas is found to obey the relation PV3/2=P V^{3 / 2}= constant during an adiabatic process. If such a gas initially at a temperature TT, is compressed to half of its initial volume, then its final temperature will be :

Options

  1. A2T2 T
  2. B4T4 T
  3. C(2)1/2T(2)^{1 / 2} T
  4. D2(2)1/2T2(2)^{1 / 2} T

Correct answer

C. (2)1/2T(2)^{1 / 2} T

Step-by-step solution

Gas obeys the equation PV3/2=P V^{3 / 2}= constant Comparing with PVγ=P V^\gamma= constant γ=32\gamma=\frac{3}{2} For adiabatic process T1V1γ−1=T2V2γ−1T(V)3/2−1=T2(V2)3/2−1T2=T(2)3/2−1=(2)1/2T\begin{aligned} T_1 V_1^{\gamma-1} & =T_2 V_2^{\gamma-1} \\ T(V)^{3 / 2-1} & =T_2\left(\frac{V}{2}\right)^{3 / 2-1} \\ T_2 & =T(2)^{3 / 2-1} \\ & =(2)^{1 / 2} T\end{aligned}
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