Question 843: Thermodynamics & KTG

Question

A monoatomic ideal gas initially at temperature '$$\mathrm{T}_1$$' is enclosed in a cylinder fitted with massless, frictionless piston. By releasing the piston suddenly the gas is allowed to expand to adiabatically to a temperature '$$\mathrm{T}_2$$'. If '$$\mathrm{L}_1$$' and '$$\mathrm{L}_2$$' are the lengths of the gas columns before and after expansion respectively, then $$\frac{\mathrm{T}_2}{\mathrm{~T}_1}$$ is

Accepted Answer

$$\begin{array}{ll} 
& 	ext { For an adiabatic process } \
& \mathrm{T}_1 \mathrm{~V}_1^{\gamma-1}=\mathrm{T}_2 \mathrm{~V}_2^{\gamma-1} \
	herefore \quad & \frac{\mathrm{T}_2}{\mathrm{~T}_1}=\left(\frac{\mathrm{V}_1}{\mathrm{~V}_2}ight)^{\gamma-1}
\end{array}$$
For a monoatomic gas, $$r=\frac{5}{3}$$
$$\begin{array}{cl} 
& \Rightarrow \gamma-1=\frac{5}{3}-1=\frac{2}{3} \
& V_1=A L_1 	ext { and } V_2=A L_2 \
	herefore \quad & \frac{T_2}{T_1}=\left[\frac{A L_1}{A L_2}ight]^{2 / 3}=\left[\frac{L_1}{L_2}ight]^{2 / 3} \
& \gamma-1=\frac{2}{3}
\end{array}$$
For an adiabatic process,
$$\begin{aligned}
& \frac{\mathrm{T}_2}{\mathrm{~T}_1}=\left(\frac{\mathrm{V}_1}{\mathrm{~V}_2}ight)^{\gamma-1}=\left(\frac{\mathrm{V}_1}{\mathrm{~V}_2}ight)^{\frac{2}{3}} \
& \mathrm{~V} \propto \mathrm{L} \
& \frac{\mathrm{T}_2}{\mathrm{~T}_1}=\left(\frac{\mathrm{L}_1}{\mathrm{~L}_2}ight)^{\frac{2}{3}}
\end{aligned}$$

Answer

$$\frac{\mathrm{L}_1}{\mathrm{~L}_2}$$

Answer

$$\frac{\mathrm{L}_2}{\mathrm{~L}_1}$$

Answer

$$\left(\frac{\mathrm{L}_1}{\mathrm{~L}_2}\right)^{2 / 3}$$

Answer

$$\left(\frac{\mathrm{L}_2}{\mathrm{~L}_1}\right)^{2 / 3}$$

Thermodynamics & KTG

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