An ideal gas at atmospheric pressure is adiab | vedclass

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(B) For an adiabatic process,the relationship between pressure $P$ and density $\rho$ is given by $P \propto \rho^{\gamma}$.Given that the initial pre

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$1.4$

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(B) For an adiabatic process,the relationship between pressure $P$ and density $ho$ is given by $P \propto ho^{\gamma}$.Given that the initial pressure $P_1 = 1 	ext{ atm}$ and the final pressure $P_2 = 128 	ext{ atm}$.The density changes from $ho$ to $ho' = 32ho$.Using the relation $\frac{P_2}{P_1} = \left(\frac{ho_2}{ho_1}ight)^{\gamma}$,we substitute the values:$\frac{128}{1} = (32)^{\gamma}$.We know that $128 = 2^7$ and $32 = 2^5$.So,$2^7 = (2^5)^{\gamma} = 2^{5\gamma}$.Equating the exponents: $7 = 5\gamma$.Therefore,$\gamma = \frac{7}{5} = 1.4$.

An ideal gas at atmospheric pressure is adiabatically compressed so that its density becomes $32$ times its initial value. If the final pressure of the gas is $128$ atmospheres,the value of $\gamma$ for the gas is:

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