Answer:
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.
