CameraIcon

SearchIcon

MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Third Law of Thermodynamics

Calculate the...

Question

Calculate the entropy increase in the evaporation of 1 mole of a liquid when it boils at $100^{0} C$ having heat of vapourisation at $100^{0} C$ as 540 Cal/gm.

Open in App

Solution

Heat of vapourisation at $100^{0} C = 540$ Cal/gm
Molar heat of vapourisation $= 540 \times 18$ Cal/gm
$Δ S = \frac{q_{r e v}}{T} = \frac{540 \times 18}{373}$
$Δ S = 26.06$ Cal/gm

Suggest Corrections

thumbs-up

0

Similar questions

The latent heat of vapourisation of water is 9700 Cal/mole and if the b.p. is 100⁰C, ebullioscopic constant of water is

Q. 4 gms of steam at

100^{0} C

is added to 20 gms of water at

46^{0} C

in a container of negligible mass. Assuming no heat is lost to surrounding, the mass of water in the container at thermal equilibrium is____.Latent heat of vaporisation = 540 cal/gm. Specific heat of water = 1 cal/gm-

^{0} C

.

Q. Standard enthalpy of vapourisation

Δ_{v a p} H^{0}

100^{0} C

40.66 k J . m o l^{- 1}

. The internal energy change of vapourisation of water at

100^{0} C

(i n k J . m o l^{- 1})

Q. Calculate the entropy change when

36 g

of liquid water evaporate at

373 K

(latent heat of vaporization of water is

0.63 k J m o l^{- 1}

Q. Correct the following statements.

a) Water boils at

100^{0} C

under atmospheric pressure.
b) The liquid evaporates above its boiling point.
c) Solids have the largest interparticle space.
d) Gases have the strongest interparticle forces.

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Third Law of Thermodynamics

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Third Law of Thermodynamics

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon