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A closed vessel contains a mixture of tw...

A closed vessel contains a mixture of two diatomic gases A and B. Molar mass of A is 16 times that of B and mass of gas A. contained in the vessel is 2 times that of B . Then

A

average kinetic energy per molecule of gas A is equal to that of gas B.

B

root mean square value of translational velocity of gas B is four times that of A.

C

root mean square value of translational velocity of gas B is four times that of A.

D

number of molecules of gas B in the cylinder is eight times that of gas A.

Text Solution

Verified by Experts

The correct Answer is:
A, B, C, D
Average KE per molecule = ` 5/2 kT `
It depends only on temperature.
` v_(rmsA) = sqrt((3RT)/(M_(A)))`
` v_(rms B) = sqrt((3RT)/(M_(B)) = sqrt((3RT)/(M_(A)//16)) = 4 v_(rms)A `
` n_(A) = m_(A)/(M_(A) ,n_(B) = m_(B)/(M_(B) = (M_(A)//2)/((M_(A)//16)) = 8 nA `
` P_(B) = ( n_(B)/(n_(A) + n_(B))) P_(0) , P_(A) = ( n_(A)/(n_(A)+ n_(B))) P_(0)`
` Rightarrow P_(B) = 8 P_(A)`
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