Question 819: Thermodynamics & KTG

Question

The pressure and density of a diatomic gas $$\left(\gamma=\frac{7}{5}\right)$$ changes adiabatically from $$(\mathrm{P}, \rho)$$ to $$\left(\mathrm{P}^{\prime}, \rho^{\prime}\right)$$. If $$\frac{\rho^{\prime}}{\rho}=32$$ then $$\frac{\mathrm{P}^{\prime}}{\mathrm{P}}$$ should be

Accepted Answer

For adiabatic process, $$\mathrm{PV}^\gamma=$$ a constant
$$	herefore \quad \frac{\mathrm{P}^{\prime}}{\mathrm{P}}=\left(\frac{\mathrm{V}^{\prime}}{\mathrm{V}}ight)^\gamma=\left(\frac{ho^{\prime}}{ho}ight)^\gamma$$
Given that, $$\frac{ho^{\prime}}{ho}=32$$ and $$\gamma=\frac{7}{5}$$
$$	herefore \quad \frac{\mathrm{P}^{\prime}}{\mathrm{P}}=(32)^{\frac{7}{5}}=128$$

Answer

$$\frac{1}{128}$$

Answer

128

Answer

32

Answer

64

Thermodynamics & KTG

The pressure and density of a diatomic gas $\left(\gamma=\frac{7}{5}\right) changes adiabatically from (\mathrm{P}, \rho) to \left(\mathrm{P}^{\prime}, \rho^{\prime}\right). If \frac{\rho^{\prime}}{\rho}=32 then \frac{\mathrm{P}^{\prime}}{\mathrm{P}}$ should be

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