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Two vessels A and B contain ideal gases ...

Two vessels A and B contain ideal gases with the temperature of B double that of A. Both gases are heated, so that they attain the same temperature. It is found that the fractional increase in the most probable speed of gas in vessel A is double that of the mean speed of gas in B. The ratio of the final to the initial temperature of gas in vessel A is

A

`3 - 2(sqrt(2))`

B

`2 - 3(sqrt(2))`

C

`3 + 2(sqrt(2))`

D

`2 + 3(sqrt(2))`

Text Solution

Verified by Experts

The correct Answer is:
C

`E_(A) = 3/2 kT_(A), E_(B) = 3/2k (2T_A)`
and `E_(A)^(') = 3/2 kT_(A) , E_(B)^(') = 3/2 kT`
`(Delta v_(P))/(v_P) = (sqrt(T/(T_A))-1) = 2 (Delta v_(avg))/(v_(avg)) = 2 (sqrt((T)/(2T_(A)))-1)`
Let, `T//T_(A) = x` the above equation reduces to
`(sqrt(x)-1) = 2(sqrt(x//2)-1) or, sqrt(x)(sqrt(2)-1)=1`
or, `x = [1//(3-2sqrt(2))]`
or, `x = 3+2sqrt(2)`.
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