Question

The average velocity of CO2 at T K is 9×10^4 cm/s. The value of T is?

Answer 1

Answer:

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Answer 2

Answer:

T=1428915.083 K.

Explanation:

Since we know the kinetic theory of gas

According to this theory, we know that the K.E.=$\frac{mv^{2} }{2}$...............(1)

we also know the average K.E. of the gas=$\frac{3kT}{2}$..............(2)

where k=boltzmann constant=$\frac{R}{N_{A} }$

So, average K.E. of the gas=$\frac{3RT}{2N_{A} }$......................(3)

From (1) and (3) we get,

$v= \sqrt{\frac{3RT}{M} }$................(4)

As, No. of moles =Given mass(m)/Molar mass(M)

Molar mass of carbon dioxide=12+16x2=44

$v=9\times10^{2}$m/s

so,$(9\times 10^{2}) ^{2} =\frac{3\times 8.314\times T}{44}$

So, T=1428915.083 K.