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(B) The total pressure inside the vessel is equal to the external pressure,which is $P_{total} = 1 \ atm$.The unknown compound exerts a vapour pressur

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(B) The total pressure inside the vessel is equal to the external pressure,which is $P_{total} = 1 \ atm$.The unknown compound exerts a vapour pressure of $0.68 \ atm$. Since it is in equilibrium with its vapour,the partial pressure of the unknown compound is $P_{unknown} = 0.68 \ atm$.The partial pressure of $He$ is $P_{He} = P_{total} - P_{unknown} = 1 \ atm - 0.68 \ atm = 0.32 \ atm$.Using the ideal gas law for $He$: $PV = nRT$.$0.32 \ atm 	imes V = 0.1 \ mol 	imes 0.0821 \ L \cdot atm \cdot K^{-1} \cdot mol^{-1} 	imes 273 \ K$.$0.32 	imes V = 2.24133$.$V = \frac{2.24133}{0.32} \approx 7.004 \ L$.Thus,the total volume is close to $7 \ L$.

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