Question

Calculate w and u for the conversion of one mole of water at 100 c to steam at 1atm pressure. Heat of vaporization of water at 100 c is 40670 j.Mol1

Answer 1

Answer:

Volume of 0.5 mole of steam at 1 atm pressure

=

P

nRT

=

1.0

0.5×0.0821×373

=15.3L

Change in volume = Vol. of steam − Vol. of water

=15.3− negligible =15.3L

Work done by the system,

w=P

ext

×volumechange

=1×15.3=15.3litre−atm

=15.3×101.3J=1549.89J

w should be negative as the work has been done by the system on the surroundings

w=−1549.89J

Heat required to convert 0.5 mole of water at 100

o

C to steam

=0.5×40670J=20335J

According to the first law of thermodynamics

ΔU=q+w=20335−1549.89=18785.11J

Answer 2

Given:

Heat of vaporization of water at 100 c is 40670 j/mol.

To find:

The work done and Internal Energy change in this thermodynamic process.

Calculation:

During vapourisation , there is no change in temperature ;the heat given is used to change the state from liquid to gas.

So , internal energy change :

$\rm{ \therefore \: \Delta U = \mu \times C_{v} \times \Delta \theta}$

$\rm{ = > \: \Delta U = \mu \times C_{v} \times 0}$

$\boxed{ \rm{ = > \: \Delta U = 0 \: joule}}$

Now , applying 1st Law of Thermodynamics:

$\rm{ \therefore \: \Delta Q = W + \Delta U }$

$\rm{ = > \: \Delta Q = W + 0}$

$\rm{ = > \: \Delta Q = W }$

$\boxed{ \rm{ = > \: W = 40670 \: joules }}$

Hope It Helps.