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At the instant the boiling point is reached, the vapour pressure of the solution should be equal to atm. pressure. And since the boiling is continued the pressure will keep on increasing. Thus the relationship between the pressures will be be $P_5 < P_{10} < P_{15}$.

And since at boiling the temperature of the liquid remains constant the temperature should be equal at all interval. That is Option C should be correct.

But the given answer is option A which doesn't make any sense to me. Can you help me with the question?

  • $\begingroup$ Boiling of a solution differs from boiling of a pure liquid a great deal. To begin with, there is no longer such thing as boiling point. $\endgroup$ Dec 4 '19 at 16:15
  • $\begingroup$ When making candy, the boiling temperature increases as water boils off and the sugar concentration in the liquid increases, see e.g. craftybaking.com/howto/candy-sugar-syrup-temperature-chart $\endgroup$ Dec 4 '19 at 16:56

As more of the solvent is boiled off, the molality of the solute increases, which increases the boiling point (as $\Delta T_B = mK_b$). Therefore, $T_5<T_{10}<T_{15}$ (Karsten Theis mentioned a good link in the comments giving a practical example)

Since the solution is boiled in an open vessel, the vapour pressure of the solvent will at all times be equal to the atmospheric pressure, as it is an isobaric process. Therefore, $P_5=P_{10}=P_{15}=P_{\text{atm}}$

  • $\begingroup$ I understand the case for temperature. However, the same can't be said for pressure. Anyways, Thanks a lot😃. $\endgroup$
    – Asad Ahmad
    Dec 4 '19 at 19:08
  • $\begingroup$ Your argument for the temperature relationship seems straight forward to me. However, I don't agree with your reasoning for the vapor pressure relationship (though the relationship itself is correct). The vapor pressure of the solution is by definition equal to atmospheric pressure, so the vapor pressure is equal to atmospheric pressure at every time and thus P5=P10=P15=Patm. $\endgroup$
    – airhuff
    Dec 5 '19 at 0:11
  • $\begingroup$ @airhuff didnt notice the 'open vessel' statement in the question. Edited accordingly. $\endgroup$ Dec 5 '19 at 8:29

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