Air is filled in a motor tube at 27^{circ}C a | vedclass

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(C) When the tube suddenly bursts,the process is adiabatic.For an adiabatic process,the relationship between temperature and pressure is given by: $\f

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$150 K$

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(C) When the tube suddenly bursts,the process is adiabatic.For an adiabatic process,the relationship between temperature and pressure is given by: $\frac{T_2}{T_1} = \left(\frac{P_2}{P_1}ight)^{\frac{\gamma - 1}{\gamma}}$.Here,the initial temperature $T_1 = 27^{\circ}C = 27 + 273 = 300 K$.The initial pressure $P_1 = 8 	ext{ atm}$.After the tube bursts,the final pressure $P_2 = 1 	ext{ atm}$ (atmospheric pressure).The adiabatic index $\gamma = 1.5$.Substituting the values: $\frac{T_2}{300} = \left(\frac{1}{8}ight)^{\frac{1.5 - 1}{1.5}} = \left(\frac{1}{8}ight)^{\frac{0.5}{1.5}} = \left(\frac{1}{8}ight)^{\frac{1}{3}}$.Since $8 = 2^3$,we have $\left(\frac{1}{8}ight)^{\frac{1}{3}} = \frac{1}{2}$.Therefore,$T_2 = 300 	imes \frac{1}{2} = 150 K$.

Air is filled in a motor tube at $27^{\circ}C$ and at a pressure of $8$ atmospheres. The tube suddenly bursts,then the temperature of the air is $[\text{Given } \gamma \text{ of air} = 1.5]$.

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