Question

show that monoatomic gas the ratio of two specific heat is 3:5​

Answer 1

E=3/2N(A) k(B) T

:. and molar specific heat at constant volume

C(V) = dE/dT =3/2 N(A) k(B)=3/2R

:. C(p) - C(v) =R using this eqn

:. C(p) = 5/2

:. gama = Cp/Cv = 5/3

Explanation:

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