Question

Initially, 1.00 mol of an ideal monatomic gas has 75.0 J of thermal energy. Part A If this energy is increased by 22.0 J , what is the change in entropy?

Answer 1

Δ S = 2.51 × 10²³ J/K

Step-by-step explanation:

change of entropy formula

`\Delta S = (3)/(2)Nln(E_x)/(E_n)`

N = 6.02 × 10²³ atoms

Initial thermal energy = Eₙ = 75 J

Final thermal energy,

Eₓ = 75 J + 22 J

Eₓ = 97 J

`\Delta S = (3)/(2)Nln(E_x)/(E_n)`

`\Delta S = (3)/(2)* 6.02* 10^(23) ln(97)/(75)`

Δ S = 2.51 × 10²³ J/K