Find ratio of quantities of water and of argon in airtight container I am working on my EJU exam past papers (2013) 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.



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
$$\begin{align}n_{g,1}&amp;=n_{g,H_2O}+n_{Ar}\\n_{H_2O}&amp;=n_{g,H_2O}+n_{l,H_2O}\\n_{g,2}&amp;=n_{H_2O}+n_{Ar}\end{align}$$
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.