Question 867: 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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