The pressure of a moist gas at 27^{circ}C is | vedclass

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(C) The total pressure of a moist gas is the sum of the partial pressure of the dry gas and the vapour pressure of water: $P_{total} = P_{gas} + P_{H_

Vedclass · Accepted Answer

$22$

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(C) The total pressure of a moist gas is the sum of the partial pressure of the dry gas and the vapour pressure of water: $P_{total} = P_{gas} + P_{H_2O}$.Given $P_{total} = 4 \ atm$ and $P_{H_2O} = 0.4 \ atm$,so $P_{gas} = 4 - 0.4 = 3.6 \ atm$.According to Boyle's Law,at constant temperature,$P_1V_1 = P_2V_2$. When the volume is doubled $(V_2 = 2V_1)$,the new partial pressure of the dry gas becomes $P_{gas, new} = P_{gas} / 2 = 3.6 / 2 = 1.8 \ atm$.The vapour pressure of water remains constant at a given temperature,so $P_{H_2O} = 0.4 \ atm$.The new total pressure is $P_{total, new} = 1.8 + 0.4 = 2.2 \ atm$.Since $2.2 \ atm = 22 	imes 10^{-1} \ atm$,the value of $x$ is $22$.

Column-$I$	Column-$II$
$A$. Hydrogen gas $(P=200 \ atm, T=273 \ K)$	$p$. Compressibility factor $Z \neq 1$
$B$. Hydrogen gas $(P \approx 0, T=273 \ K)$	$q$. Attractive forces are dominant
$C$. $CO_2$ $(P=1 \ atm, T=273 \ K)$	$r$. $PV=nRT$
$D$. Real gas with very large molar volume	$s$. $P(V_m-b)=RT$

The pressure of a moist gas at $27^{\circ}C$ is $4 \ atm$. The volume of the container is doubled at the same temperature. The new pressure of the moist gas is $x \times 10^{-1} \ atm$. Find the value of $x$. (Nearest integer) (Given: The vapour pressure of water at $27^{\circ}C$ is $0.4 \ atm$)

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