The pressure in the tyre of a car is four times the atmospheric pressure at 300 K. If this tyre suddenly bursts, its new temperature will be (γ=1.4)

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300 (4)1.4/0.4

300 14−0.4/1.4

300 (2)−0.4/1.4

300 (4)−0.4/1.4

answer is D.

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Detailed Solution

For adiabatic process TγPγ−1=constant⇒T2T1=P1P21−γγ⇒T2300=41(1−1.4)1.4⇒T2=300(4)−0.41.4

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