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Calculate the pressure exerted by $10^{23}$ gas molecules each of mass $10^{-22}$ g in a container of volume one litre. The rms speed is $10^5 cm \;sec^{-1}$

$\begin{array}{1,1}(a)\;3\times10^7dyne\;cm^{-2}\\(b)\;3.3\times10^4dyne\;cm^{-2}\\(c)\;3.3\times10^7dyne\;cm^{-2}\\(d)\;none\;of\;these \end {array}$

1 Answer

Given
n = $10^{23}$
$m = 10^{-22} g$
V = 1 litre = $10^3 cm^3$
$U_{rms} = 10^5 cm sec^{-1}$
PV = $\large\frac{1}{3} mn{u^2_{rms}}$
$P\times10^3 = \large\frac{1}{3}\times10^{-22}\times10^{23}\times(10^5)^2$
$P = 3.3\times10^7 dyne \;cm^{-2}$
Hence answer is (c)
answered Feb 15, 2014 by sharmaaparna1
 

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