Be sure to answer all parts.A sample of natural gas contains 6.816 moles of methane (CH4), 0.589 moles of ethane (C2H6), and 0.381 moles of propane (C3H8). If the total pressure of the gases is 3.93 atm, what are the partial pressures of the gases?

Respuesta :

Answer:

Explanation:

Given parameters:

number of moles of CH₄  = 6.816 moles

number of moles of C₂H₆  = 0.589 mole

number of moles of C₃H₈  =  0.381 moles

Total pressure of the gases  = 3.93 atm

Unknown:

Partial pressure of the gases = ?

Solution:

Dalton's law of partial pressure states that " the total pressure of a mixture of gases is equal to the sum of the partial pressures of the constituent gases".

                         P[tex]_{T}[/tex]   = P[tex]_{1}[/tex]  + P[tex]_{2}[/tex]  + P[tex]_{3}[/tex]...............

where P[tex]_{T}[/tex]  = total pressure of the gas mixture

           P₁, P₂, P₃......  = partial pressure of gas 1, 2, 3............

The partial pressure of a gas is the pressure it would exert if confined alone in the volume of the gas mixture.

      Partial pressure of gas  = mole fraction of gas x total pressure of mixture

Now let us solve the problem.

Total number of moles of gases  = 6.816 moles + 0.589 mole + 0.381 moles

                                                        = 7.786moles

Partial pressure of CH₄ = [tex]\frac{6.816}{7.786} x 3.93 = 3.44 atm[/tex]

Partial pressure of C₂H₆  = [tex]\frac{0.589}{7.786}[/tex]  x 3.93  = 0.30atm

Partial pressure of C₃H₈  = [tex]\frac{0.381}{7.786}[/tex] x 3.93  = 0.19atm