Answer:
internal energy E 3716.35 j
cv = 12.47 J/K
S = 12.47 J/K
A = 0.29 J
[tex]\mu =3716.35 J/mole[/tex]
Explanation:
given data:
Kr atomic number = 36
degree of freedom = 3
1) internal energy E = [tex]\frac{f}{2} n R T[/tex]
= [tex]\frac{3}{2} *1*8.314*298 = 3716.35 j[/tex]
2) [tex]cv = \frac{E}{T}[/tex]
[tex]= \frac{3716.35}{298} = 12.47 J/K[/tex]
3) [tex]S = cv =\frac{E}{T} = 12.47 J/K[/tex]
4) A, Halmholtz free energy = E -TS = 37146.35 - 12.47*298 = 0.29 J
5)[tex]chemcial potential \mu = \frac{energy}{mole} = 3716.35 J/mole[/tex]