Chemistry: Help with Partial Pressure Calculation

A 712.0 mL cylinder containing SF6 at a pressure of 0.100 atm is connected by a valve to 0.763 L cylinder containing C5H10 at 342.0 torr pressure. Calculate the partial pressure (torr) of C5H10 when the valve is opened.

To calculate the partial pressure of C5H10 after the valve is opened, we need to know what the mole fraction of C5H10 is. We can calculate moles of each gas (before opening the valve) using the ideal gas law. Assume a temperature of 25 C = 298 K (the T doesn't really matter when calculating a mole fraction).

For SF6:

PV = nRT
n = PV/RT = (0.100 atm)(0.712 L) / (0.0821 L atm / K mole)(298 K) = 0.00291 moles SF6

For C5H10: NOTE: 342.0 torr x (760 torr / 1 atm) = 0.450 atm

n = PV/RT = (0.450 atm)(0.763 L) / (0.0821 L atm / K mole)(298 K) = 0.0140 moles C5H10

total gas moles = moles SF6 + moles C5H10 = 0.00291 + 0.0140 = 0.0169 moles

mole fraction C5H10 = (moles C5H10 / total gas moles) = 0.0140 / 0.0169 = 0.828

The total gas pressure after opening the valve is found using the ideal gas law. The total volume is now equal to 0.712 L + 0.763 L = 1.475 L.

P = nRT/V = (0.0169 moles)(0.0821 L atm / K mole)(298 K) / (1.475 L) = 0.280 atm

Partial pressure C5H10 = Total pressure x mole fraction C5H10 = (0.280 atm)(0.828) = 0.232 atm

The easiest way to calculate the answer is to completely ignore the SF6 in the tank, since you only need to consider the C5H10.

When you open the valve, the C5H10 will expand into a larger volume, and so its pressure will be:

P1V1 = P2V2
342 Torr (0.763L) = P2 (1.475L)
P2 = 177 torr