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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Thermodynamics
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A monatomic ideal gas at temperature $ T_1$ is enclosed in a cylinder with frictionless piston. The gas is allowed to expand adiabatically to a temperature $T_2$ by releasing piston suddenly. If $L_1$ and $L_2$ are lengths of gas columns before and after expansion, then

$\begin {array} {1 1} (A)\;\large\frac{T_1}{T_2} = \bigg( \large\frac{L_1}{L_2} \bigg)^{ \large\frac{2}{3}} & \quad (B)\;\large\frac{T_2}{T_1} = \bigg( \large\frac{L_1}{L_2} \bigg)^{ \large\frac{5}{3}} \\ (C)\;\large\frac{T_1}{T_2} = \bigg( \large\frac{L_2}{L_1} \bigg)^{ \large\frac{5}{3}} & \quad (D)\;\large\frac{T_1}{T_2} = \bigg( \large\frac{L_2}{L_1} \bigg)^{ \large\frac{2}{3}} \end {array}$

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For adiabatic process, $T_1V_1^{\gamma -1} = T_2V_2^{\gamma-1}$ if $‘a’$ is cross-section of vessel then
$V_1 = a.L_1 \: and \: V_2 = a.L_2$
So, $ \large\frac{T_1}{T_2}$$ = \bigg( \large\frac{V_2}{V_1} \bigg)^{\gamma-1}$$ = \bigg( \large\frac{aL_2}{aL_1} \bigg)^{\gamma-1}$$ = \bigg( \large\frac{L_2}{L_1} \bigg)^{\gamma-1}$
Now for monatomic gases, $ \gamma \bigg(=\large\frac{C_p}{C_v} \bigg) $$= \large\frac{5}{3}$
So, $ \large\frac{T_1}{T_2}$$ = \bigg( \large\frac{L_2}{L_1} \bigg)^{\large\frac{5}{3}-1}$$ = \bigg (\large\frac{L_2}{L_1} \bigg)^{\large\frac{2}{3}}$
Ans : (D)
answered Mar 20, 2014 by thanvigandhi_1
edited Mar 23, 2014 by balaji.thirumalai
 

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