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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Kinetic Theory of Gases
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RMS speed of a monoatomic gas is increased by 2 times.If the process is done adiabatically then the ratio of initial volumes to final volume will be

$\begin{array}{1 1}(A)\;4\\(B)\;4^{2/3}\\(C)\;2^{3/2}\\(D)\;8\end{array} $

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1 Answer

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$\large\frac{V_{RMS1}}{V_{RMS2}}=\sqrt{\large\frac{T_1}{T_2}}$
$\Rightarrow 4T_1=T_2$
For an adiabatic process
$TV^{r-1}\Rightarrow$ Constant
$T_1V_1^{2/3}=T_2V_2^{2/3}$
$\large\frac{V_1}{V_2}$$=(4)^{3/2}=8$
Hence (D) is the correct answer.
answered May 12, 2014 by sreemathi.v
 

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