In an adiabatic process,the density of a diat | vedclass

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(D) For an adiabatic process,the relation between pressure $P$ and volume $V$ is given by $PV^{\gamma} = 	ext{constant}$.Since density $ho = \frac{

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$128$

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(D) For an adiabatic process,the relation between pressure $P$ and volume $V$ is given by $PV^{\gamma} = 	ext{constant}$.Since density $ho = \frac{m}{V}$,we have $V = \frac{m}{ho}$.Substituting this into the adiabatic equation: $P \left(\frac{m}{ho}ight)^{\gamma} = 	ext{constant}$.Since mass $m$ is constant,we get $P \propto ho^{\gamma}$.Therefore,$\frac{P_f}{P_i} = \left(\frac{ho_f}{ho_i}ight)^{\gamma}$.For a diatomic gas,the adiabatic index $\gamma = \frac{7}{5} = 1.4$.Given $ho_f = 32 ho_i$,we have $\frac{ho_f}{ho_i} = 32$.Thus,$n = \frac{P_f}{P_i} = (32)^{7/5} = (2^5)^{7/5} = 2^7 = 128$.

In an adiabatic process,the density of a diatomic gas becomes $32$ times its initial value. The final pressure of the gas is found to be $n$ times the initial pressure. The value of $n$ is

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