3. (physics) In a vessel, there is a mixture of equal masses of nitrogen $N_{2}$ and helium He at a pressure $p$. The absolute temperature of the gas is doubled, during which all nitrogen molecules dissociate into atoms. Find the pressure of the gas mixture at this temperature. The molar masses of the gases are $\mu_{\text {He }}=4$ g/mol, $\mu_{N_{2}}=28$ g/mol. The gases are considered ideal.
Solution. Let the mass of helium and nitrogen in the vessel be $m$. Then Dalton's law for the initial gas mixture gives

$$
p=\frac{v_{\mathrm{He}} R T}{V}+\frac{v_{N_{2}} R T}{V}
$$

where

$$
v_{N_{2}}=\frac{m}{\mu_{N_{2}}}=v ; \quad v_{\mathrm{He}}=\frac{m}{\mu_{\mathrm{He}}}=7 v
$$

are the amounts of substance of nitrogen and helium in the vessel, $V$ - its volume. From this, we have for the initial pressure of the mixture

$$
p=\frac{8 v R T}{V}
$$

After heating, the vessel will contain a mixture of atomic nitrogen and helium. The amount of substance of atomic nitrogen after dissociation doubles, the amount of substance of helium remains unchanged

$$
v_{N}^{\prime}=2 v ; \quad v_{H e}^{\prime}=7 v
$$

Therefore, the pressure in the vessel will be equal to

$$
p^{\prime}=\frac{v_{H e}^{\prime} R 2 T}{V}+\frac{v_{N}^{\prime} R 2 T}{V}=\frac{7 v R 2 T}{V}+\frac{2 v R 2 T}{V}=\frac{18 v R T}{V}=\frac{9}{4} p
$$