Figure shows two vessels with adiabatic walls, one containing 0.1 g of helium `(gamma =1.67 , M = 4 g mol^(-1))` and the other containing some amount of hydrogen `(gamma = 1.4, M = 2 g mol^(-1))`. Initially , the temperatures of the two gases are equal. The gases electrically heated for some time during which equal amounts of heat are given to the gases. It is found that the temperatures rise through the same amount in the two vessels. Calculate the mass of hydrogen.

Explanation

`m_(He) = 0.1` g,
`gamma_1 = 1.67`, `M_(He) = 4g// mol.`
`M_(H2) = ? `, `M_(H2)= 2g//mol`,
`gamma_2=1.4`
Since it is an adiabatic surroundings
For He, `dQ = nCvdT`
`=(m)/2 xx (R)/(gamma-1) xx dT`
`0.1/4 xx (R)/(1.67 -1) xx dT`
For `H_2` , `dQ = nCvdT`
`=m/2 xx (R)/(gamma-1) xx dT`
`m/2 xx (R)/(1.4-1) xx dT`
[where m is the required mass of `H_2`.] Since equal amount of heat is given to both, so, `dQ` is same in both Equations ...
(i) and ... (ii), we get
`0.1/4 xx (R)/(0.67) xx dT`
`=m/2 xx R/0.4 xx dT`
implies ` m= 0.1/2 xx (0.4)/(0.67)`
`=0.0298 = 0.03g`.