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

Answer:

7.71 atm

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:

$\frac{p}{n}=\frac{RT}{V}=const$

This means for two conditions, we'd obtain:

$\frac{p_1}{n_1}=\frac{p_2}{n_2}$

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:

$p_2 = p_1\cdot \frac{n_2}{n_1}$

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:

$p_2 = p_1\cdot \frac{n_2}{n_1}$

And:

$\chi_{H_2}=\frac{n_H_2}{n_2}$

The equation becomes:

$p_{H_2}=\chi_{H_2}p_2=p_1\cdot \frac{n_2}{n_1}\cdot \frac{n_H_2}{n_2}=p_1\cdot \frac{n_H_2}{n_1}$

Substituting the variables:

$p_{H_2}=9 atm\cdot \frac{4 mol + 2 mol}{7 mol}=7.71 atm$