Find ratio of quantities of water and of argon in airtight container

I am working on my [EJU exam past papers (2013)](https://www.jasso.go.jp/en/eju/examinee/pastpaper_sample/pastpaper_2013_2.html) and got stuck at this question. Knowledge of high school or first year university chemistry is assumed. I honestly do not know where I should start. 



> 
> When argon and water were placed in an airtight container with a
>  constant volume and the gas mixture was kept at 373 K, the pressure of
>  the gas mixture became P1 with a portion of water remained as liquid.
>  When the whole was heated to 403 K, all the water was vaporized and
>  the pressure became P2. From (1) to (6) below choose the correct
>  formula representing the ratio, water/argon, of the quantities (in
>  mol) of water and of argon in the container. Assume that the vapor
>  pressure of water at 373 K is Pv, that argon is an ideal gas not
>  dissolving in water, and that the volume of water before heating to
>  403 K can be neglected.
> 
> 
> 


[![](https://i.stack.imgur.com/6amlB.png)](https://i.stack.imgur.com/6amlB.png)
Please note that answer key provided is (1). I tried to think that we could form an equation from total moles of gases equate to the sum of total moles of water and argon at 403 K since all are vaporized. Then use the ideal gas law to find the total amount of gases at 403 K. However, I cannot connect these pieces and information given at 373 K together. Could anyone give me a solution to this question or at least a hint so I can continue working on it?




You should use the ideal gas law to convert between $n\_i$ (amount of gas) and $p/T$. For each temperature and pressure pair you can write
$$\begin{align}n\_{g,1}=\frac{p\_1V}{RT\_1}\\n\_{g,2}=\frac{p\_2V}{RT\_2}\end{align}$$
where
$$\begin{align}n\_{g,1}&=n\_{g,H\_2O}+n\_{Ar}\\n\_{H\_2O}&=n\_{g,H\_2O}+n\_{l,H\_2O}\\n\_{g,2}&=n\_{H\_2O}+n\_{Ar}\end{align}$$
You also know that for each individual gas the partial pressure is proportional to the number of moles (Dalton's law):
$$n\_i=\frac{p\_iV}{RT}$$
 The "trick" is that at $T\_1=\pu{373K}$
$$p\_{Ar} = p\_1 - p\_V$$
where $p\_V$ is given as the partial pressure of water at that temperature.
You should be able to show that 
$$\frac{n\_{H\_2O}}{n\_{Ar}}=\frac{n\_{g,2}}{n\_{Ar}}-1$$
and making appropriate substitutions it should not be difficult to arrive at the answer.


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