1A )
using
TV^{\gamma -1}=constant
When one mole of monoatomic gas at T K undergoes adiabatic change under a constant external pressure of 1 atm changes volume from 1 L to 2 L.The fianl temperature in Kelvin would be
A. \frac{T}{2^{2/3}}
B. T + 2/3 * 0.0821
C. T
D. T - 2/3 * 0.0821
-
UP 0 DOWN 0 0 4
4 Answers
1FOR ADIA BATIC PROCESS......PVγ=CONSTANT
TVγ-1=K
MONOATOMIC GASES HAVE γ=5/3
T*1=T2*22/3
T2=T/22/3.........(A)
106guys ... look at the wording of the question...
think its a bit ambiguous...
under a constant external pressure this is also there....
so i dont think (quoting BT solution in my own way)... that we can apply PVγ = c
11Ans given by fiitjee is also A(like anirudh and avinav),,but i was having a doubt.