5 points total) The latent heat of vaporization (boiling) of helium at a pressure of 1 atm and a temperature of 4.2 K is 21.8 kJ kg-1 . The densities at 4.2 K of the liquid and vapor are 125 kg m-3 and 19 kg m-3 respectively. a. (2 points) What proportion of latent heat is involved in work against the interatomic attraction? b. (2 points) Estimate the depth e of the potential well which results from the forces between two helium atoms. Give your answer in Joules, electron-volts (eV) and kelvin (K), where the answer in kelvin corresponds to an energy kT. (Take the average number of nearest neighbors to be z = 10 and note that the atomic number of helium is 4.) c. (1 point) Why does helium turn into a gas at such low temperatures?

a) proportion of latent heat involved in work against the interatomic attraction = 0.794

b)Depth of the well in Joules = 23 * 10^-24 J

ii) In eV, E = 0.000144 eV

III) in Kelvin, E = 1.67 K

C) Check the explanation section for C

Explanation:

a) Latent heat, Q = 21.8 kJ/kg

Vapor density, Vd = 19 kg/m^3

Liquid density, Ld = 125 kg/ m^3

Pressure, P = 1 atm = 1 * 10^5 Pa

Volume change from liquid to vapor = (1/Vd) - (1/ Ld)

Volume change = (1/19) - (1/125)

Volume change = 0.045 m^3

Work done in converting from liquid to vapor, W = P * (Volume change)

W = 1 *0.045 * 10^5

W = 4.5 kJ

Let the proportion of latent heat involved in work against the interatomic attraction be Pr

Pr = (Q - W)/Q

Pr = (21.8 - 4.5)/21.8

Pr = 0.794

b) To calculate the depth of the potential well :

I) In joules

n(L-W) = 0.5 z Na E

z = 10

Where E = depth of the well

4(21.8-4.5) = 0.5 * 10 * 6.02 * 10^23 * E

E = 23 * 10^-24 J

ii) In eV

E = ( 23 * 10^-24)/(1.6 * 10^-19)

E = 0.000144 eV

III) In Kelvin

E = ( 23 * 10^-24)/(1.38 * 10^-23)

E = 1.67 K

C) Helium turns to a gas at that low temperature because of the large workdone (4.5 kJ) against the interatomic attraction.

a and b sometimes have a positive value