Relation between P, V, T, Gamma in Adiabatic Proceses
When 1 mole o...
Question
When 1 mole of a mono atomic ideal gas a T K undergoes an adiabatic change under a constant external pressure of 1 atm, its volume changes from 1L to 2L. The final temperature in Kelvin is:
A
T22/3
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B
T+23×0.0821
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C
T
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D
T−23×0.0821
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Solution
The correct option is DT−23×0.0821 In an adiabatic process, Q = 0
From first law of thermodynamics, ΔU=Q+W ΔU=W...1
Now W = −P(V2−V1) (given Pext=1atm
= -1(2-1) atm L = -1 atm L
For calculating ΔU we need to calculate Cv Cv = Rγ−1
for monoatomic gas γ=53
Hence Cv=3R2
Now using eq. 1 ΔU=nCvdT
Now using eq. 1 1×3R2(Tf−Ti)=−1atmL
Assuming final temperature as Tf and solving for it we get Tf=Ti−23R