Question

a 5L container contains 3 moles of helium and 4 moles of hydrogen at a pressure of 9 atms maintaining a constant T and additional 2 mol or hydrogen are added.... what is the new partial pressure of hydrogen gas in a container

Answer 1

7.71 atm

Step-by-step explanation:

Given the following data:

`V = 5 L`

`n_(He) = 3 mol`

`n_(H_2) = 4 mol`

`p_1 = 9 atm`

`T = const`

According to the ideal gas law, we know that the product between pressure and volume of a gas is equal to the product between moles, the ideal gas law constant and the absolute temperature:

`pV = nRT`

Since the temperature and the ideal gas constant are constants, as well as the fixed container volume of 5 L, we may rearrange the equation as:

`(p)/(n)=(RT)/(V)=const`

This means for two conditions, we'd obtain:

`(p_1)/(n_1)=(p_2)/(n_2)`

Given:

`p_1 = 9 atm`

`n_1 = n_(initial total) = n_(He) + n_(H_2) = 3 mol + 4 mol = 7 mol`

`n_2 = n_(final total) = n_(He) + n_(H_2) = 3 mol + 4 mol + 2 mol = 9 mol`

Solve for the final pressure:

`p_2 = p_1\cdot (n_2)/(n_1)`

Now, according to the Dalton's law of partial pressures, the partial pressure is equal to the total pressure multiplied by the mole fraction of a component:

`p_(H_2)=\chi_(H_2)p_2`

Knowing that:

`p_2 = p_1\cdot (n_2)/(n_1)`

And:

`\chi_(H_2)=(n_H_2)/(n_2)`

The equation becomes:

`p_(H_2)=\chi_(H_2)p_2=p_1\cdot (n_2)/(n_1)\cdot (n_H_2)/(n_2)=p_1\cdot (n_H_2)/(n_1)`

Substituting the variables:

`p_(H_2)=9 atm\cdot (4 mol + 2 mol)/(7 mol)=7.71 atm`