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
Answers
Answered by
14
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
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
rajeshhooda:
How u got ans i m getting 44RT please explain
Answered by
58
The answer is 16RT
As the He is monoatomic therefore the internal energy of He is equal to
=> 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
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