Question

The pressure inside a tyre is 4 times that of atmosphere. If the tyre bursts suddenly at temperature 300 K, what will be the new temperature ?

Difficult

Solution

Under adiabatic change
$\frac{T_{2}}{T_{1}} = {(\frac{P_{1}}{P_{2}})}^{\frac{1 - γ}{γ}} or T_{2} = T_{1} {(P_{1} / P_{2})}^{\frac{1 - γ}{γ}} ∴ T_{2} = 300 (4 / 1)^{\frac{1 - (7 / 5)}{(7 / 5)}} (γ = 1 \cdot 4 = 7 / 5 for air) or T_{2} = 300 (4)^{- 2 / 7}$

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