Question

A mixture of nitrogen and hydrogen gases, at a total pressure of 663 mm Hg, contains 3.46 grams of nitrogen and 0.156 grams of hydrogen. What is the partial pressure of each gas in the mixture? PN2 = mm Hg PH2 = mm Hg

Answer 1

Answer: The partial pressure of nitrogen gas is 405.76 mmHg and that of hydrogen gas is 257.24 mmHg

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

• For nitrogen gas:

Given mass of nitrogen gas = 3.46 g

Molar mass of nitrogen gas = 28 g/mol

Putting values in above equation, we get:

$\text{Moles of nitrogen gas}=\frac{3.46g}{28g/mol}=0.123mol$

• For hydrogen gas:

Given mass of hydrogen gas = 0.156 g

Molar mass of hydrogen gas = 2 g/mol

Putting values in above equation, we get:

$\text{Moles of hydrogen gas}=\frac{0.156g}{2g/mol}=0.078mol$

Mole fraction of a gas is calculated by using the formula:

$\chi_{A}=\frac{n_{A}}{n_{A}+n_{B}}$      ......(1)

• For nitrogen gas:

Putting values in equation 1, we get:

$\chi_{\text{nitrogen gas}}=\frac{0.123}{0.123+0.078}=0.612$

• For hydrogen gas:

Putting values in equation 1, we get:

$\chi_{\text{hydrogen gas}}=\frac{0.078}{0.123+0.078}=0.388$

The partial pressure of a gas is given by Raoult's law, which is:

$p_A=p_T\times \chi_A$     ......(2)

where,

$p_A$ = partial pressure of substance A

$p_T$ = total pressure  = 663 mmHg

$\chi_A$ = mole fraction of substance A

• For nitrogen gas:

$p_{\text{Nitrogen gas}}=663mmHg\times 0.612\\\\p_{\text{Nitrogen gas}}=405.76mmHg$

• For hydrogen gas:

$p_{\text{Hydrogen gas}}=663mmHg\times 0.388\\\\p_{\text{Hydrogen gas}}=257.24mmHg$

Hence, the partial pressure of nitrogen gas is 405.76 mmHg and that of hydrogen gas is 257.24 mmHg