Question

A gas mixture contains 320mg methane, 175 mg argon, 225 mg nitrogen (N2). The partial pressure of argon at 300K is 12.52 kPa. What is the volume and total pressure of the mixture?

Answer 1

Answer: The volume and total pressure of the mixture is 0.876 L and 89.36 kPa respectively.

Explanation:

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

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$     .....(1)

• For methane:

Given mass of methane = 320 mg = 0.3 g     (Conversion factor:  1 g = 1000 mg)

Molar mass of methane = 16 g/mol

Putting values in equation 1, we get:

$\text{Moles of methane}=\frac{0.3g}{16g/mol}=0.019mol$

• For argon:

Given mass of argon = 175 mg = 0.175 g

Molar mass of argon = 40 g/mol

Putting values in equation 1, we get:

$\text{Moles of argon}=\frac{0.175g}{40g/mol}=0.0044mol$

• For nitrogen:

Given mass of nitrogen = 225 mg = 0.225 g

Molar mass of nitrogen = 28 g/mol

Putting values in equation 1, we get:

$\text{Moles of nitrogen}=\frac{0.225g}{28g/mol}=0.0080mol$

To calculate the volume of the mixture, we use the equation:

PV = nRT         ......(2)

We are given:

Partial pressure of argon = 12.52 kPa

Temperature = 300 K

R = Gas constant = $8.31\text{L kPa }mol^{-1}K^{-1}$

n = number of moles of argon = 0.0044 moles

Putting values in equation 2, we get:

$12.52kPa\times V=0.0044mol\times 8.31\text{L kPa }mol^{-1}K^{-1}\times 300K\\\\V=\frac{0.0044\times 8.31\times 300}{12.52}=0.876L$

Now, calculating the total pressure of the mixture by using equation 2:

Total number of moles = [0.019 + 0.0044 + 0.0080] mol = 0.0314 mol

V= volume of the mixture = 0.876 L

Putting values in equation 2, we get:

$P\times 0.876L=0.0314mol\times 8.31\text{L kPa }mol^{-1}K^{-1}\times 300K\\\\P=\frac{0.0341\times 8.31\times 300}{0.876}=89.36kPa$

Hence, the volume and total pressure of the mixture is 0.876 L and 89.36 kPa respectively.