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 505 mmHg. (R=0.082 L-atm/K mol)
For equal number of mols of each, then
Xpropane = 0.5 and
Xbutane = 0.5.
Then partial p propane = 0.5*505
To calculate the partial pressure of propane in the mixture, we need to use Dalton's law of partial pressures. According to Dalton's law, the total pressure is the sum of the partial pressures of each component in the mixture.
Let's start by calculating the partial pressure of propane. Since the mixture contains equal numbers of moles of propane and butane, we can assume that the mole fraction of propane is 0.5 (50%).
First, let's convert the temperature to Kelvin. 20 °C is equivalent to 20 + 273.15 = 293.15 K.
To calculate the partial pressure, we can use the following equation:
Partial pressure of propane = mole fraction of propane * total pressure
Given:
Mole fraction of propane (Xpropane) = 0.5
Total pressure (Ptotal) = 505 mmHg
Gas constant (R) = 0.082 L-atm/K mol
Partial pressure of propane = Xpropane * Ptotal
= 0.5 * 505 mmHg
To convert mmHg to atm, divide by 760:
Partial pressure of propane = 0.5 * 505 mmHg / 760
= 0.3334 atm (rounded to four decimal places)
Therefore, the partial pressure of propane in the mixture is approximately 0.3334 atm at 20 °C and 505 mmHg.