Answer all the questions and attach to your lab report Name: Date 1. (a) caj Ske

Question

Answer all the questions and attach to your lab report Name: Date 1. (a) caj Sketch a graph to show how the pressure of hexane (Cathz, a volatile liquid) varies as a etch a graph to show how the pressure of hexane (CsH12, a volatile liquid) varies as a function of time at constant temperature. Ans.: volatile liquid was placed in a container at 25 "C and the pressure measured. if the tneris now exposed to a vacuum, would the pressure be the same, higher, or lower if (b) A (i) some liquid remains Ans.: (i) no liquid remains Ans.: Toluene (C7H8) has a normal boiling point of 110.6 °C. What is the boiling point at a pressure of 15 mm. 2. AHvap 38.6 kJ/mol. R 8.314 J-Kmol1. Ans.:

Explanation / Answer

1.
a) Pressure for volatile liquids vary slightly more than proportional to temperature. So, with respect to time, the graph should be a constant graph. Had i been for temperature, the graph would have been somewhat proportional to temperature.

b) Whenever we try to create vaccum inside the container having volatile liquid, we are trying to create some volume inside by some means for e.g. pulling the piston and fixing the volume to some extent so as to create vaccum. So this vaccum has created some vapour pressure apart from the liquid present. Also we should notice that the liquid must have decreased to some extent.

i) When some liquid is present, the vaccum must have created some pressure inside which indicates some gaseous substance that arose from the liquid. So, the pressure increases. But the pressure increases to an extent and becomes constant afterwards. By this we can conclude that all the liquid is not converted to vapour instantaneously. They are in a condition known as Phase equilibrium.

ii) Even if we try to create more volume inside the container by creating more vaccum, we are still going to get that same equilibrium pressure as far as there is some liquid present. When there is not liquid present, then all the liquid is vapourised and they increase the vapour pressure.

2.

By Clausius-Clapeyron Equation:
ln(P2/P1) = Hvap/R * (1/T2 - 1/T1)

Hvap = 38.6 k/mol
R = 8.314 J/mol/K
P1= 760 mm, as it is given as normal configuration
T1 = 110.6 C = 383.6 C
P2 = 15 mm
T2 = ?

By putting the values in the equation, I got T2 as 288.56 K = 15.56 C.
The boiling should decrease according to the equation as the pressure is decreasing.

Let me know if you have any doubts.

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