Question

A mixture containing 0.477 mol he(g), 0.265 mol ne(g), and 0.115 mol ar(g) is confined in a 7.00-l vessel at 25 ∘c. part a calculate the partial pressure of he in the mixture.

Answer 1

Q1)
we can use the ideal gas law equation to find the total pressure of the system ;
PV = nRT
where P - pressure
V - volume - 7 x 10⁻³ m³
n - number of moles 
total number of moles - 0.477 + 0.265 + 0.115 = 0.857 mol
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature in K - 273 + 25 °C = 298 K
substituting the values in the equation 
 P x 7 x 10⁻³ m³ = 0.857 mol x 8.314 Jmol⁻¹K⁻¹ x 298 K
P = 303.33 kPa
1 atm = 101.325 kPa
Therefore total pressure - 303.33 kPa / 101.325 kPa/atm = 2.99 atm

Q2)
partial pressure is the pressure exerted by the individual gases in the mixture.
partial pressure for each gas can be calculated by multiplying the total pressure by mole fraction of the individual gas.

total number of moles - 0.477 + 0.265 + 0.115 = 0.857 mol
mole fraction of He - $\frac{0.477}{0.857} = 0.557$
mole fraction of Ne - $\frac{0.265}{0.857} = 0.309$
mole fraction of Ar - $\frac{0.115}{0.857} = 0.134$
partial pressure - total pressure x mole fraction
partial pressure of He - 2.99 atm x 0.557 = 1.67 atm
partial pressure of Ne - 2.99 atm x 0.309 = 0.924 atm
partial pressure of Ar - 2.99 atm x 0.134 = 0.401 atm