Question

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

Answer 1

$m{H}_e=0.1g,$

${\gamma }_1=1.67,M_{H}e=4g/mol.,$

$M_{H}2=?;M_{H}2=2g/mol,$

${\gamma }_2=1.4$

Since it is an adiabatic surroundings

For He, dQ = nCvdT

= $\frac{m}{2}\times \frac{R}{\gamma -1}\times dT$

= $\frac{0.1}{4}\times \frac{R}{(1.67-1)}\times dT$

For $H_2,dQ=nCvdT$

= $\frac{m}{2}\times \frac{R}{\gamma -1}\times dT$

= $\frac{m}{2}\times \frac{R}{(1.4-1)}\times dT$

[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

$\frac{0.1}{4}\times \frac{R}{0.67}\times dT$

= $\frac{m}{2}\times \frac{R}{0.4}\times dT$

$\Rightarrow m=\frac{0.1}{2}\times \frac{0.4}{0.67}$

= 0.0298 = 0.03g.