(4)

According to Charle’s law, when the pressure of a gas is kept constant, the volume of the gas is directly proportional to the temperature of the gas.

i.e.,  $V\propto T\Rightarrow \frac{V}{T}=\text{Constant}$

Therefore, the plot of V vs T is a straight line passing through the origin.

So, $P_1>P_2>P_3$

(4)

Using ideal gas equation for mixture of two gases,

$P=\frac{({\mu }_1+{\mu }_2)RT}{V}$

$=\frac{(0.2+0.3)\times 8.3\times 200}{200\times {10}^{-6}}=\frac{0.5\times 8.3\times 200}{200\times {10}^{-6}}$

$P=4.15\times {10}^6$ $\text{Pa}$

(4)

$\mu =\frac{\text{Mass of water}}{\text{Molecular weight of water}}$

           $=\frac{4.5 \text{kg}}{18\times {10}^{-3} \text{kg}}=250$

At STP, $T=273$ $\text{K}$ and $P={10}^5$ ${\text{N/m}}^2$

From, $PV=\mu RT\Rightarrow V=\frac{\mu RT}{P}$

$V=\frac{250\times 8.3\times 273}{{10}^5}$

$V=5.66$ ${\text{m}}^3$

(2)

As $PV=nRT$

or  $n=\frac{PV}{RT}=\frac{\text{mass}}{\text{molar mass}}$                       ...(i)

$\text{Density, }\rho =\frac{\text{mass}}{\text{volume}}=\frac{\left(\text{molar mass}\right)P}{RT}=\frac{\left(m{N}_A\right)P}{RT}$

$\therefore$   $\rho =\frac{mP}{kT}$                                         $(\because R=N_{A}k)$

(4)

According to an ideal gas equation, the molecular weight of an ideal gas is

            $M=\frac{\rho RT}{P}$                              $(\text{as }P=\frac{\rho RT}{M})$

where $P,T$ and $\rho$ are the pressure, temperature and density of the gas respectively, and $R$ is the universal gas constant.

$\therefore$     The molecular weight of A is

          $M_A=\frac{{\rho }_{A}R{T}_A}{P_A} \text{and that of B is} M_B=\frac{{\rho }_{B}R{T}_B}{P_B}$

Hence, their corresponding ratio is

        $\frac{M_A}{M_B}=\left(\frac{{\displaystyle {\rho }_A}}{{\displaystyle {\rho }_B}}\right)\left(\frac{{\displaystyle T_A}}{{\displaystyle T_B}}\right)\left(\frac{{\displaystyle P_B}}{{\displaystyle P_A}}\right)$

Here, $\frac{{\rho }_A}{{\rho }_B}=1.5=\frac{3}{2}, \frac{T_A}{T_B}=1 \text{and} \frac{P_A}{P_B}=2$

$\therefore$    $\frac{M_A}{M_B}=\left(\frac{{\displaystyle 3}}{{\displaystyle 2}}\right)\left(1\right)\left(\frac{{\displaystyle 1}}{{\displaystyle 2}}\right)=\frac{3}{4}$