Question

A flask contains 0.25 mole of SO2(g), 0.50 mole of CH4(g), and 0.50 mole of O2(g). The total pressure of the gases in the flask is 800 mm Hg. What is the partial pressure of the SO2(g) in the flask?

Answer 1

Answer: The partial pressure of sulfur dioxide gas in the flask is 160 mmHg

Explanation:

We are given:

Moles of sulfur dioxide gas = 0.25 moles

Moles of methane gas = 0.50 moles

Moles of oxygen gas = 0.50 moles

To calculate the mole fraction of sulfur dioxide, we use the equation:

$\chi_{SO_2}=\frac{n_{SO_2}}{n_{SO_2}+n_{CH_4}+n_{O_2}}\\\\\chi_{SO_2}=\frac{0.25}{0.25+0.50+0.50}=0.2$

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

$p_{SO_2}=p_T\times \chi_{SO_2}$

where,

$p_{SO_2}$ = partial pressure of sulfur dioxide gas

$p_T$ = total pressure = 800 mmHg

$\chi_{SO_2}$ = mole fraction of sulfur dioxide = 0.2

Putting values in above equation, we get:

$p_{SO_2}=800\times 0.2=160mmHg$

Hence, the partial pressure of sulfur dioxide gas in the flask is 160 mmHg