Question

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?

Answer 1

a. About 83% of the latent heat is involved in work against the interatomic attraction. b. The depth of the potential well is approximately 4.62 x 10^-23 J, 2.88 x 10^14 eV, and 33.6 K. c. Helium turns into a gas at low temperatures due to its weak interatomic attraction.

Step-by-step explanation:

a. To calculate the proportion of latent heat involved in work against interatomic attraction, we need to first calculate the work done using the formula W = P(V2 - V1), where P is the pressure and V2 and V1 are the volumes of the vapor and liquid respectively. The difference in volume is V2 - V1 = 1 - 19 = -18 m^3. Since the work done is negative (work is done against the interatomic attraction), the proportion of latent heat involved in work is given by the ratio |W|/Q, where Q is the latent heat of vaporization. Therefore, the proportion is |(-1 atm)(-18 m^3)/(21.8 kJ/kg)| = 0.8292, or about 83%.

b. The depth e of the potential well can be estimated using the formula e = kT/2, where k is Boltzmann's constant and T is the temperature. The average number of nearest neighbors z is 10, and the atomic number of helium is 4. Therefore, the depth e can be calculated as e = (4 * (1.38 x 10^-23 J/K) * 4.2 K)/(2 * 10) = 4.62 x 10^-23 J. This is equivalent to approximately 2.88 x 10^14 eV and 33.6 K.

c. Helium turns into a gas at such low temperatures because its interatomic attraction is weak. The low mass and low atomic number of helium result in weak intermolecular forces, making it easier for helium atoms to overcome the attractive forces and transition from a liquid to a gas state at low temperatures.

Answer 2

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

Step-by-step 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.