A tyre pumped to a pressure of 2 text{ atm} s | vedclass

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(B) The process of a tyre bursting is an adiabatic expansion process.For an adiabatic process, the relation between temperature $T$ and pressure $P$ i

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$\left(\frac{1}{2}\right)^{1/3} T$

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(B) The process of a tyre bursting is an adiabatic expansion process.For an adiabatic process, the relation between temperature $T$ and pressure $P$ is given by $T^\gamma P^{1-\gamma} = 	ext{constant}$.Thus, $T_1^\gamma P_1^{1-\gamma} = T_2^\gamma P_2^{1-\gamma}$.Given: $P_1 = 2 	ext{ atm}$, $P_2 = 1 	ext{ atm}$ (atmospheric pressure), $T_1 = T$, and $\gamma = \frac{3}{2}$.Substituting these values:$T^{\frac{3}{2}} (2)^{1 - \frac{3}{2}} = T_2^{\frac{3}{2}} (1)^{1 - \frac{3}{2}}$$T^{\frac{3}{2}} (2)^{-\frac{1}{2}} = T_2^{\frac{3}{2}}$$T_2 = T 	imes (2)^{-\frac{1}{2} 	imes \frac{2}{3}}$$T_2 = T 	imes (2)^{-\frac{1}{3}}$$T_2 = \left(\frac{1}{2}ight)^{1/3} T$.

$A$ tyre pumped to a pressure of $2 \text{ atm}$ suddenly bursts. If the temperature of air before expansion is $T$, then the air temperature after the tyre bursts is (Assume the expansion is adiabatic and adiabatic constant $\gamma = \frac{3}{2}$).

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