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

1.	Show that monoatomic gas the ratio of two specific heat is 3:5
Answer» 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: hope it's HELP full

Answer»

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:

hope it's HELP full

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

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