Question

A mixture of ideal gas has 2 moles of He, 4 moles of oxygen and 1 mole of ozone at absolute temperature T. The internal energy of the mixture is

a. 13RT b. 11RT c. 16RT d. 14RT

Answer 1

U=n1f12RT+n2f22RT

f - No.of degrees of freedom

f1 for a diatomic gas = 5

f2 for a monoatomic gas =

Therefore , U=2×52RT+4×32RT

=11RT


therefore answer is b =11RT

Answer 2

The answer is 16RT

As the He is monoatomic therefore the internal energy of He is equal to

$u = \frac{f}{2} nrt$

=> U=3/2 nRT

as the O2 is diatomic therefore

=>U=5/2nRT

and the last one O3 is triatomic and non-linear

so U=6/2nRT

put the value of n=moles

and add all

you will get the answer as 16RT