A mixture of one mole each of $$H_2$$, He and $$O_2$$ each are enclosed in a cylinder of volume V at temperature T. If the partial pressure of $$H_2$$ is 2 atm, the total pressure of the gases in the cylinder is:

First, we recall the ideal-gas equation for any single component: $$P\,V = n\,R\,T$$ where $$P$$ is the pressure exerted by that component, $$V$$ is the volume it occupies, $$n$$ is the amount of substance in moles, $$R$$ is the universal gas constant, and $$T$$ is the absolute temperature.

Because the gases behave ideally and are contained in the same vessel, each gas exerts its own pressure as if the others were absent. This individual pressure is called the partial pressure. Mathematically, for any one gas, we have $$P_i = \dfrac{n_i R T}{V}.$$

We are told that one mole of $$H_2$$ exerts a partial pressure of $$2\;\text{atm}$$. Substituting $$n_{H_2}=1$$ into the formula gives

$$P_{H_2} \;=\; \dfrac{(1)\,R\,T}{V} \;=\; 2\;\text{atm}.$$

From this equality we can express the common factor $$\dfrac{R\,T}{V}$$ directly:

$$\dfrac{R\,T}{V} \;=\; 2\;\text{atm}.$$

Now, the mixture contains three different gases—$$H_2$$, He, and $$O_2$$—each present in one mole. Hence the total amount of substance is

$$n_{\text{total}} = 1 + 1 + 1 = 3\;\text{mol}.$$

The total pressure of the mixture, according to Dalton’s law, is the sum of the partial pressures. Using the ideal-gas expression for the whole mixture, we have

$$P_{\text{total}} = \dfrac{n_{\text{total}}\,R\,T}{V}.$$

Substituting $$n_{\text{total}} = 3$$ and our earlier result $$\dfrac{R\,T}{V} = 2\;\text{atm}$$, we obtain

$$P_{\text{total}} = 3 \times 2\;\text{atm} = 6\;\text{atm}.$$

Hence, the correct answer is Option A.