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One mole of monoatomic ideal gas T(K) undergoes reversible adiabatic change under a constant external pressure of 1 atm. Changes volume from 1 litre to 2 litre. The final temperature is:
  1. T+23×R
  2. T
  3. T32R
  4. T22/3

A
T+23×R
B
T32R
C
T22/3
D
T
Solution
Verified by Toppr

For adiabatic condition, dq=0
Thus by first law of thermodynamics.
dU=dW+dq
dU=dW(1) as dq=0 for adiabatic
dW=Pext(V2V1)
dU=CvdT
For monoatomic gas
Degree of freedom= 3M=3×1=3 All transitional
Cv=32K or for 1 mole K=R
Cv=32R
Putting all values in equation 1, we have
CvdT=Pext(V2V1)
32(T2T1)=1(2L1L)
(T2T1)=1×23R
T2=T123R
Thus, final temperature is T23R .

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