State the no. Of freedom possessed by a mono atomic molecule in space. Also give the expression for total energy possessed by it at a given temperature. Hence give the total energy of the atom at 300k.
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Mono atomic gas atom :
Degree of freedom = 3A - R
3x1 -0 = 3
as we know f = 3
so total energy per molecule = 3/2 K(B) T
K(x) = 1/2 K(B)T
K(y) = 1/2 K(B)T
K(z) = 1/2 K(B)T
Given T = 300 K
so Total energy = 3/2 K(B)T = 3/2 x 1.38 x 10⁻²³ x 300
= 621 x 10⁻²³
= 6.21 x 10⁻²¹ joules
Degree of freedom = 3A - R
3x1 -0 = 3
as we know f = 3
so total energy per molecule = 3/2 K(B) T
K(x) = 1/2 K(B)T
K(y) = 1/2 K(B)T
K(z) = 1/2 K(B)T
Given T = 300 K
so Total energy = 3/2 K(B)T = 3/2 x 1.38 x 10⁻²³ x 300
= 621 x 10⁻²³
= 6.21 x 10⁻²¹ joules
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