Air is filled in a tube of the wheel of a car at 27∘C and 2 atm pressure if the tube is suddenly bursts, the final temperature of air will be- (γ=1.5,21/3=1.251)
−33∘C
0∘C
21.6∘C
240∘C
A
0∘C
B
240∘C
C
−33∘C
D
21.6∘C
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Solution
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Given,
T1=27+273=300K
P1=2atm
γ=1.521/3=1.251
In the process
Tγ1Pγ−11=Tγ2Pγ−12
(T1T2)γ=(P2P1)γ−1
(300T2)3/2=(21)1/2
T2=30021/3=2400C
The correct option is D.
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