Question 922: Thermodynamics & KTG

Question

A polyatomic gas $$(\gamma=4 / 3)$$ is compressed to $$\left(\frac{1}{8}ight)^{	ext {th }}$$ of its volume adiabatically. If its initial pressure is $$\mathrm{P}_0$$, its new pressure will be

Accepted Answer

For an adiabatic process, the following relation holds:
$$P_1V_1^\gamma = P_2V_2^\gamma$$
where:
$$P_1$$ is the initial pressure
$$V_1$$ is the initial volume
$$P_2$$ is the final pressure
$$V_2$$ is the final volume
$$\gamma$$ is the adiabatic index
In this case, we have:
$$P_1 = P_0$$
$$V_1 = V$$
$$V_2 = \frac{V}{8}$$
$$\gamma = \frac{4}{3}$$
Substituting these values into the adiabatic equation, we get:
$$P_0V^{\frac{4}{3}} = P_2 \left(\frac{V}{8}\right)^{\frac{4}{3}}$$
Simplifying the equation:
$$P_0V^{\frac{4}{3}} = P_2 \left(\frac{V^{\frac{4}{3}}}{8^{\frac{4}{3}}}\right)$$
$$P_0V^{\frac{4}{3}} = P_2 \left(\frac{V^{\frac{4}{3}}}{16}\right)$$
$$P_2 = 16P_0$$
Therefore, the new pressure is $$16P_0$$.
So the answer is
Option D: $$16P_0$$
.

Answer

$$2 \mathrm{P}_0$$

Answer

$$8 \mathrm{P}_0$$

Answer

$$6 \mathrm{P}_0$$

Answer

$$16 \mathrm{P}_0$$

Thermodynamics & KTG

A polyatomic gas $(\gamma=4 / 3) is compressed to \left(\frac{1}{8}\right)^{\text {th }} of its volume adiabatically. If its initial pressure is \mathrm{P}_0$, its new pressure will be

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