(D) For an adiabatic process, the relationship between pressure $P$ and density $\rho$ (or $d$) is given by $P \propto \rho^\gamma$. Thus, $\frac{P^{\

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(D) For an adiabatic process, the relationship between pressure $P$ and density $\rho$ (or $d$) is given by $P \propto \rho^\gamma$. Thus, $\frac{P^{\prime}}{P} = \left(\frac{d^{\prime}}{d}\right)^\gamma$. Given $\frac{d^{\prime}}{d} = 32$ and $\gamma = \frac{7}{5}$. Substituting these values: $\frac{P^{\prime}}{P} = (32)^{7/5}$. Since $32 = 2^5$, we have $\frac{P^{\prime}}{P} = (2^5)^{7/5} = 2^7$. Calculating $2^7 = 128$. Therefore, the ratio $\frac{P^{\prime}}{P} = 128$.

Class 11 Physics Thermodynamics Adiabatic Process

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The pressure and density of a given mass of a diatomic gas $\left(\gamma = \frac{7}{5}\right)$ change adiabatically from $(P, d)$ to $(P^{\prime}, d^{\prime})$. If $\frac{d^{\prime}}{d} = 32$, then $\frac{P^{\prime}}{P}$ is $(\gamma = \text{ratio of specific heats})$.

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A
$\frac{1}{128}$
B
$\frac{1}{64}$
C
$64$
D
$128$

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The pressure and density of a given mass of a diatomic gas $\left(\gamma = \frac{7}{5}\right)$ change adiabatically from $(P, d)$ to $(P^{\prime}, d^{\prime})$. If $\frac{d^{\prime}}{d} = 32$, then $\frac{P^{\prime}}{P}$ is $(\gamma = \text{ratio of specific heats})$.

Similar Questions

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