Question

A cylinder of indefinite height contains a frictionless and massless piston. The part of the cylinder...

A cylinder of indefinite height contains a frictionless and massless piston. The

part of the cylinder above the piston is exposed to the atmosphere with P=1.000 atm.

In the bottom of the cylinder are 5.000 kmoles of a liquid that has a boiling point

equal to 320.0 K at P=1.000 atm., a heat of vaporization of 2.520 X 107J/kmole, and

cp = a+bT-1  for the liquid phase, where a=3.000x104 J/kmole/K and b=2.500x104J/kmole.


A. At constant pressure, how much heat is required to convert all 5 kmoles of

this liquid, initially at 29.00K and 1.000 atm, to an ideal gas or vapor at 320.0

K within the cylinder? Recall that at the boiling point, the vapor pressure of a gas is equal to the

surrounding atmospheric pressure. Work in kilojoules.


B. If the volume of the liquid at the boiling point is 0.0500 m3

per kmole, how much of the latent

heat of vaporization goes into changing the internal energy of the vapor? Neglect the expansion of

the liquid as it is heated to the boiling point.

Homework Answers

Answer #1

A. first the liquid will heat upto 320k and then conert to gas at the same temperature.

The process shall be isobaric as the pressure will be constant and the piston is massless, hence,

throughout the process, P=1.000atm

Now, Q=Heat transferred during the processof heating the liquid from 29K to 320K =    \(n\int_{29}^{320}(a+bT-1) dT\) , where n= no. of moles ( since dQ=n Cp dT, when the pressure is held constant)

On putting the apt. values, Q=13656.339 KJ

while the liquid is converted to gas, the amount of heat supplied = heat of vaporization=n X molar heat of vaporozation=

2.520*107*5/1000=1.3482 kJ

so total Q=1.3482+13656.339=13657.68 kJ

B. latent heat of vaporization=1.3482 kJ

it is important to note that

      

Q=1.3482 kJ as found out above

for W, we have,

dW=PdV

therefore,W=PV, where P=1.000atm and V=Volume of the gas at 320 K and 1.000 atm

from ideal gas equation, PV=nRT, we have,

on inserting apt values, T= 320K , V=nRT/P

therefore, V=131.291 cubic metre

So W= P*change in volume

W=101.325*1000*(131.291-0.0500)=13298.05 kJ

here, Q<W, this means that there might be an error in reporting the latent heat of vaporization as change in internal energy cannot be negative.

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