Partial pressure of gases in a container with a piston

Question

I've been having problems with the following problem.

Figure 1.

A container with a piston like that shown in Figure 1 is filled with $0.10$ $mol$ of Ar and $1.00$ $mol$ of water (liquid and vapor). The temperature of the entire container is $87 °C$ and the total pressure is $1$ $bar$. It is assumed that the volume of the liquid or cup is negligible with respect to that of the gas, and that the temperature of water and Ar is always maintained at $87 °C$. In this initial situation, the partial pressure of Ar is $A$ $bar$, the partial pressure of water vapor is $B$ $bar$, the volume of the gas is $C$, and the number of moles of liquid water is $D$ $mol$.

The piston is then pulled back quickly to fix the gas volume to $15.8$ $L$. At the moment when the piston is pulled, the partial pressure of Ar decreases to $E$ $bar$ and the partial pressure of water vapor decreases to $F$ $bar$. In this situation, the water will boil, because the total pressure of the gas is lower than the saturated vapor pressure of water at $87 °C$, which is $0.6226$ $bar$. The partial pressure of water vapor increases due to the boiling until the boiling eventually stops. When the boiling stops, the number of moles of liquid water is $G$ $mol$. After that, evaporation proceeds until equilibrium is reached. At equilibrium, the number of moles of liquid water is $H$ $mol$.

Calculate the appropriate numerical values for blanks $A$ - $H$.

The reason I am having trouble solving the problem is the fact that there seems to be too many unknown variables: the initial volume of the container $(V_0)$, the partial pressures of Ar $(p_{Ar,1})$ and water $(p_w)$ and the number of moles of liquid water $(n_{w,l})$. I've come up with the following formulas but I am unable to get any further.

$\ce{p_{total}=p_{Ar,1} + p_{water}=1 bar}$

$\ce{n_{total}=n_{Ar,1} + n_{water,g}=0.10 mol + (1 mol-n_{water,l})}$

$\ce{p_{Ar,1}=\frac{n_{Ar}*R*T}{V_0}}$

$\ce{p_{w,1}=\frac{n_{water}*R*T}{V_0}}$

$\ce{V_0=\frac{n_{total}*R*T}{p_{total}}}$

Answer 1

$B=\pu{0.6226 bar}$ as the water-water vapour system would be in equilibrium.

Now,

$$ \begin{align} A+B&= \pu{1 bar}\\ A&= \pu{0.3774 bar} \end{align} $$

For initial volume of the container,

$$ \begin{align} P_{\ce{Ar}_i}V_i&=n_{\ce{Ar}_i}RT\\ V_i&=\frac{n_{\ce{Ar}_i}RT}{P_{\ce{Ar}_i}}\\ C&=\pu{\frac{0.1 \times 0.08314 \times 360}{0.3774} L}\\ C&=\pu{7.93 L}\\ \end{align} $$

For initial moles of water vapour,

$$ \begin{align} P_{\ce{H2O{(v)}}_i}V_i&=n_{\ce{H2O{(v)}}_i}RT\\ n_{\ce{H2O{(v)}}_i}&=\frac{P_{\ce{H2O{(v)}}_i}V_i}{RT}\\ n_{\ce{H2O{(v)}}_i}&=\pu{\frac{0.6226 \times 7.93}{360 \times 0.08314} L}\\ n_{\ce{H2O{(v)}}_i}&=\pu{0.165 mol}\\ \end{align} $$

Now,

$$ \begin{align} n_{\ce{H2O{(v)}}_i} + n_{\ce{H2O{(l)}}_i} &=\pu{1 mol}\\ n_{\ce{H2O{(l)}}_i} &= (1 - n_{\ce{H2O{(v)}}_i}) \pu{mol}\\ D &= \pu{ (1 - 0.165) mol}\\ D &= \pu{ 9.835 mol}\\ \end{align} $$

Now as the volume is doubled by Boyle's law we can say that the partial pressure of argon and water vapour will decrease to half of original.

Thus,

$$ E=\frac{A}{2}=\pu{0.1887 bar}\\ F=\frac{B}{2}=\pu{0.3113 bar}\\ $$

Now as the partial pressure of water vapour is less than the vapour pressure of water at that temperature, water will evaporate to make it equal.

Now as the temperature is same thus the vapour pressure of water is also same as initial. Thus only change in system is it's volume which was doubled. So we can conclude that twice amount of water vapour as compared to initial conditions will be required to attain this pressure.

Thus,

$$ \begin{align} n_{\ce{H2O{(v)}}_f}&=2 \times n_{\ce{H2O{(v)}}_i}\\ n_{\ce{H2O{(v)}}_f}&=\pu{0.33 mol}\\ \end{align} $$

But as the total moles of water is conserved,

$$ \begin{align} G=H=&(1-n_{\ce{H2O{(v)}}_f}) \pu{mol}\\ G=H=&\pu{0.67 mol}\\ \end{align} $$