What is the density of one mole of He (molar | vedclass

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(C) From the ideal gas equation,$PV = nRT$. Since $n = \frac{m}{M}$,we have $PV = \frac{m}{M} RT$. Rearranging for density $d = \frac{m}{V}$,we get $d

Vedclass · Accepted Answer

$1.33 	imes 10^{-1} \ g \ L^{-1}$

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(C) From the ideal gas equation,$PV = nRT$. Since $n = \frac{m}{M}$,we have $PV = \frac{m}{M} RT$. Rearranging for density $d = \frac{m}{V}$,we get $d = \frac{PM}{RT}$. Given: $P = 0.82 \ atm$,$M = 4 \ g \ mol^{-1}$,$R = 0.082 \ L \ atm \ mol^{-1} \ K^{-1}$,and $T = 300 \ K$. Substituting the values: $d = \frac{0.82 	imes 4}{0.082 	imes 300} = \frac{4}{30} = 0.1333 \ g \ L^{-1} = 1.33 	imes 10^{-1} \ g \ L^{-1}$.

What is the density of one mole of $He$ (molar mass $= 4 \ g \ mol^{-1}$) at $300 \ K$ and a pressure of $0.82 \ atm$? $(R = 0.082 \ L \ atm \ mol^{-1} \ K^{-1})$

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