Respuesta :
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:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(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:
[tex]\text{Moles of methane}=\frac{0.3g}{16g/mol}=0.019mol[/tex]
- 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:
[tex]\text{Moles of argon}=\frac{0.175g}{40g/mol}=0.0044mol[/tex]
- 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:
[tex]\text{Moles of nitrogen}=\frac{0.225g}{28g/mol}=0.0080mol[/tex]
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 = [tex]8.31\text{L kPa }mol^{-1}K^{-1}[/tex]
n = number of moles of argon = 0.0044 moles
Putting values in equation 2, we get:
[tex]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[/tex]
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:
[tex]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[/tex]
Hence, the volume and total pressure of the mixture is 0.876 L and 89.36 kPa respectively.
