Question

What is the change in entropy of helium gas with total mass 0.140 kg at the normal boiling point of helium when it all condenses isothermally to liquid helium? Assume that the normal boiling point of helium is 4.216 K and the heat of vaporization of helium is 2.09×104 J/kg.

Answer 1

Answer:694J/K

Step-by-step explanation:

m = 0.140 kg

dt = 4.216K

Entropy change,

dg = dh - dt*ds = 0

dh = dtds

ds = dh/dt

dh =m x c

s2-s1 = 0.140 x 2.09 x 10^4 /4.216

s2-s1 = 2926/4.216

s2-s1 = 694J/K

Answer 2

-0.695 kJ/K

Step-by-step explanation:

Given data

• Mass of He (m): 0.140 kg

• Normal boiling point of He (T): 4.216 K

• Heat of vaporization of He (ΔHvap): 2.09 × 10⁴ J/kg

The heat (Q) released in the condensation of 0.140 kg of He can be calculated using the following expression.

Q = - ΔHvap × m = -2.09 × 10⁴ J/kg × 0.140 kg = -2.93 × 10³ J = -2.93 kJ

The change in the entropy (ΔS) for this process is:

ΔS = Q/T = -2.93 kJ/4.216 K = -0.695 kJ/K