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(C) For an adiabatic process,the relation between pressure $P$ and volume $V$ is $PV^{\gamma} = 	ext{constant}$.Since density $d = \frac{m}{V},$ we h

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$128$

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(C) For an adiabatic process,the relation between pressure $P$ and volume $V$ is $PV^{\gamma} = 	ext{constant}$.Since density $d = \frac{m}{V},$ we have $V = \frac{m}{d},$ which implies $V \propto \frac{1}{d}.$Substituting this into the adiabatic equation: $P \left(\frac{1}{d}ight)^{\gamma} = 	ext{constant},$ or $P \propto d^{\gamma}.$Therefore,$\frac{P'}{P} = \left(\frac{d'}{d}ight)^{\gamma}.$Given $\gamma = 7/5$ and $\frac{d'}{d} = 32,$$\frac{P'}{P} = (32)^{7/5} = (2^5)^{7/5} = 2^{5 	imes (7/5)} = 2^7.$Calculating the value,$2^7 = 128.$Thus,$\frac{P'}{P} = 128.$

The pressure and density of a diatomic gas with $\gamma = 7/5$ change adiabatically from $(P, d)$ to $(P', d').$ If $\frac{d'}{d} = 32,$ then $\frac{P'}{P}$ is

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