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224 ml of triatomic gas weights 1g at 173k and 1atm .The mass of one atom of this gas is

$\begin{array}{1 1}(a)\;8.30\times 10^{-23}g&(b)\;2.08\times 10^{-23}g\\(c)\;5.53\times 10^{-23}g&(d)\;6.24\times 10^{-23}g\end{array}$

1 Answer

$PV=nRT$
$\therefore PV=\large\frac{U}{m}$$RT$
$\Rightarrow \large\frac{1}{m}$$\times 0.0821\times 273$
$\therefore m=100$
$3N$ atoms of gas weighs 100g
$\therefore$ 1 atom of gas weighs =$\large\frac{100}{3\times 6.023\times 10^{23}}$
$\Rightarrow 5.53\times 10^{-23}g$
Hence (c) is the correct answer.
answered Oct 24, 2013 by sreemathi.v
 

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