Question

Find the change in the melting point of ice at 0°C for an increase of pressure by 1 atm. Latent heat of ice at 0°C = 80 cal/g. At melting point the volumes of 1g of water and ice are 1 cm³ and 1.091 cm³ respectively; 1 atm. = 1.013 x 10° dyne/cm².​

Answer 1

Answer:

Since heat is absorbed here at constant pressure

so , 

ΔH=1440 calories

ΔU can be calculated by using the

formula 

 ΔU=ΔH−PΔV

P=1atm, 

ΔV=0.0196L−0.180L=0.0016L

P ΔV=1∗0.0016=0.0016

Latm=0.0016∗24.217calories.

(1Latm=24.217 calories)

(1Latm=24.217 calories)So ,

ΔU=1440cal−0.0016∗24.217

cal≈1439.96calories .

There is negligible work done here.

so,

you can write  ΔH≈ΔU.