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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} $

1 Answer

$\Rightarrow 4T_1=T_2$
For an adiabatic process
$TV^{r-1}\Rightarrow$ Constant
Hence (D) is the correct answer.
answered May 12, 2014 by sreemathi.v

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