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)
partial pressure is 1/2 616mmHg
http://en.wikipedia.org/wiki/Dalton%27s_law
partial pressure= Pressuretotal*molefraction
To calculate the partial pressure of propane in the mixture, we need to use the concept of Dalton's Law of Partial Pressures.
According to Dalton's law, the total pressure of a mixture of gases is equal to the sum of the partial pressures of each individual gas.
In this case, we know that the total pressure of the mixture is 616 mmHg. Since equal numbers of moles of propane and butane are present, their partial pressures will be equal.
Let's calculate the partial pressure of propane:
1. Start by determining the number of moles of propane and butane. Since they are present in equal numbers, we'll only need to calculate the moles of one gas.
Given that the total pressure is 616 mmHg, we can use the ideal gas law equation PV = nRT, where:
- P is the pressure in atm (we need to convert mmHg to atm),
- V is the volume in liters,
- n is the number of moles,
- R is the ideal gas constant (0.082 L-atm/K mol), and
- T is the temperature in Kelvin.
2. Convert the total pressure from mmHg to atm. 1 atm = 760 mmHg, so the total pressure in atm would be 616 mmHg / 760 mmHg/atm = 0.81 atm.
3. Convert the given temperature from degrees Celsius to Kelvin. The conversion formula is K = °C + 273.15. Therefore, 20 °C + 273.15 = 293.15 K.
4. Rearrange the ideal gas law equation to solve for the number of moles (n):
n = (PV) / (RT)
5. Plug in the values into the equation:
n = (0.81 atm) / ((0.082 L-atm/K mol) × 293.15 K)
6. Calculate the number of moles. 0.81 atm / (0.082 L-atm/K mol × 293.15 K) ≈ 0.032 moles of propane (C3H8).
Since propane and butane are present in equal numbers of moles, the partial pressure of propane in the mixture is 0.81 atm × 0.5 = 0.405 atm (or 305 mmHg).