One mole of an ideal gas (Cp /Cv = γ) at absolute temperature T1 is diabatically compressed from an initial pressure P1 to a final pressure P2. The resulting temperature T2 of the gas is given by
see full answer
Your Exam Success, Personally Taken Care Of
1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya
a
T2=T1P2/P1γ/(γ−1)
b
T2=T1P2/P1(γ−1)/γ
c
T2=T1P2/P1γ
d
T2=T1P2/P1γ−1
answer is B.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
For adiabatic process, PVγ= constant For an ideal gas PV = RT or V = (RT/ P)∴ P(RT/P)γ= constant or Tγ/Pγ−1= constant (∵R= constant )∴ T2γP2γ−1=T1γP1γ−1or T2=T1P2/P1(γ−1)/γ
One mole of an ideal gas (Cp /Cv = γ) at absolute temperature T1 is diabatically compressed from an initial pressure P1 to a final pressure P2. The resulting temperature T2 of the gas is given by