Calculate the partial pressure of propane in a mixture that contains equal numbers of moles of propane (C3H8) and butane (C4H10) at 20 °C and 616 mmHg. (R=0.082 L-atm/K mol)
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To calculate the partial pressure of propane in the mixture, you need to know the total pressure and the mole fractions of propane and butane.
The mole fraction of a component in a mixture is defined as the ratio of the number of moles of that component to the total number of moles in the mixture.
In this case, the mixture contains equal numbers of moles of propane and butane. Let's assume there are 'x' moles of propane and 'x' moles of butane.
The total number of moles in the mixture is then 2x (since there are equal numbers of moles of propane and butane).
Now, we need to calculate the mole fractions of propane and butane. The mole fraction (X) of a component is given by the ratio of the number of moles of that component to the total number of moles in the mixture:
Mole fraction of propane (X_propane) = moles of propane / total moles in the mixture = x / (2x) = 1/2
Mole fraction of butane (X_butane) = moles of butane / total moles in the mixture = x / (2x) = 1/2
Since the mole fractions add up to 1 (X_propane + X_butane = 1), we know that the mole fraction of butane is also 1/2.
Now that we have the mole fraction of propane (X_propane = 1/2) and the total pressure (P_total = 616 mmHg), we can calculate the partial pressure of propane (P_partial_propane) using Dalton's Law of Partial Pressures.
According to Dalton's Law, the partial pressure of a component in a mixture is equal to the mole fraction of that component multiplied by the total pressure of the mixture.
P_partial_propane = X_propane * P_total
P_partial_propane = (1/2) * 616 mmHg
P_partial_propane = 308 mmHg
Therefore, the partial pressure of propane in the mixture is 308 mmHg.