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)
Hmmm. what did you fing wrong with this?
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To calculate the partial pressure of propane in the mixture, we can use the concept of partial pressure and the ideal gas law.
First, let's define the given information:
- The total pressure of the mixture is 616 mmHg.
- The temperature is 20 °C, which we need to convert to Kelvin (K) by adding 273.15 (20 °C + 273.15 = 293.15 K).
- The gas constant (R) is 0.082 L-atm/K mol.
Now, let's calculate the partial pressure of propane:
1. Since the mixture contains an equal number of moles of propane and butane, we can assume that the moles of propane (n) and the moles of butane (n) are the same.
2. Using the ideal gas law equation PV = nRT, we can rearrange it to solve for the number of moles (n) as follows:
n = PV / RT
Where:
- P is the total pressure of the mixture (616 mmHg or 616 torr).
- V is the volume of the gas (not given in this case but not needed for the partial pressure calculation).
- R is the gas constant (0.082 L-atm/K mol).
- T is the temperature in Kelvin (293.15 K).
3. Let's calculate the number of moles of the gas (n) by plugging in the values:
n = (616 mmHg)(1 atm/760 mmHg) / (0.082 L-atm/K mol)(293.15 K)
Here, we convert the given pressure from mmHg to atm using the conversion factor 1 atm/760 mmHg.
4. Simplify the equation:
n = 0.0813 mol
Since the number of moles of propane and butane are equal, each gas has a mole fraction of 0.5.
5. To find the partial pressure of propane, we multiply its mole fraction by the total pressure:
Partial pressure of propane = mole fraction of propane × total pressure
Partial pressure of propane = (0.5)(616 mmHg) = 308 mmHg
Therefore, the partial pressure of propane in the mixture at 20 °C is 308 mmHg.