The pressure P_{1} and density d_{1} of a dia | vedclass

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(C) For an adiabatic process,the relation between pressure $P$ and density $d$ is given by $P \propto d^{\gamma}$.Thus,$\frac{P_{2}}{P_{1}} = \left(\f

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$4$

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(C) For an adiabatic process,the relation between pressure $P$ and density $d$ is given by $P \propto d^{\gamma}$.Thus,$\frac{P_{2}}{P_{1}} = \left(\frac{d_{2}}{d_{1}}ight)^{\gamma}$.Given $\frac{d_{2}}{d_{1}} = 32$ and $\gamma = \frac{7}{5}$,we have $\frac{P_{2}}{P_{1}} = (32)^{7/5} = (2^5)^{7/5} = 2^7 = 128$.Using the ideal gas law $P = \frac{dRT}{M}$,we have $T \propto \frac{P}{d}$.Therefore,$\frac{T_{2}}{T_{1}} = \frac{P_{2}}{P_{1}} 	imes \frac{d_{1}}{d_{2}}$.Substituting the values,$\frac{T_{2}}{T_{1}} = 128 	imes \frac{1}{32} = 4$.So,the temperature becomes $4$ times its initial temperature.

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