Question 936: Thermodynamics & KTG

Question

A monoatomic gas at pressure '$$\mathrm{P}$$' having volume '$$\mathrm{V}$$' expands isothermally to a volume $$2 \mathrm{~V}$$ and then adiabatically to a volume $$16 \mathrm{~V}$$. The final pressure of the gas is $$\left(\gamma=\frac{5}{3}\right)$$

Accepted Answer

For isothermal process: $$\mathrm{P}_1 \mathrm{~V}_1=\mathrm{P}_2 \mathrm{~V}_2$$
$$	herefore \quad \mathrm{P}_2=\frac{\mathrm{P}_1 \mathrm{~V}_1}{\mathrm{~V}_2}=\frac{\mathrm{P}_1}{2} \quad\left(\because \frac{\mathrm{v}_1}{\mathrm{v}_2}=\frac{1}{2}ight)$$
For adiabatic process : $$\quad \mathrm{P}_2 \mathrm{~V}_2^\gamma=\mathrm{P}_3 \mathrm{~V}_3^\gamma$$
$$\begin{aligned}
	herefore \quad P_3 & =P_2\left(\frac{V_2}{V_3}ight)^\gamma=P_2\left(\frac{2 V_1}{V_3}ight)^\gamma=\frac{P_1}{2}\left(\frac{1}{8}ight)^{5 / 3} \quad\left[\because \frac{V_1}{V_3}=\frac{1}{16}ight] \
& =\frac{P_1}{2} \cdot \frac{1}{32}=\frac{P_1}{64}
\end{aligned}$$

Answer

$$\frac{P}{64}$$

Answer

$$\frac{\mathrm{P}}{128}$$

Answer

$$\frac{P}{8}$$

Answer

$$\frac{\mathrm{P}}{32}$$

Thermodynamics & KTG

A monoatomic gas at pressure '$\mathrm{P}' having volume '\mathrm{V}' expands isothermally to a volume 2 \mathrm{~V} and then adiabatically to a volume 16 \mathrm{~V}. The final pressure of the gas is \left(\gamma=\frac{5}{3}\right)

Choose the correct answer: