1 mole of NH3 gas at 27°C is expanded in reversible adiabatic condition to make volume 8 times (y=1.33). Final temperature and work done respectively are:
A
150 K, 909 cal
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B
150 K, 400 cal
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C
250 K, 1000 cal
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D
200 K, 800 cal
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Solution
The correct option is C 150 K, 909 cal
As we know that, for an adiabatic expansion,
TVγ−1=constant
∴T2T1=(V1V2)γ−1
Given:-
T1=27℃=(27+273)K=300K
T2=T=?
V1=V
V2=8V
γ=1.33=43
∴T300=(V8V)43−1
⇒T=3002=150K
Now, using first law of thermodynamics, ΔE=q+w
As we know that, for adiabatic expansion, q=0
∴ΔE=w
If gas expands, the internal energy decreases.
∴w=ΔE=−nCVdT
⇒w=−[1×Rγ−1×(150−300)]
⇒w=−[−150×21.33−1]
w=909cal
Hence the value of final temperature and work done are 150K and 909cal respectively.