The stopcock connecting a 2.64 L bulb containing helium gas at a pressure of 6.06 atm,

Question

The stopcock connecting a 2.64 L bulb containing helium gas at a pressure of 6.06 atm, and a 7.11 L bulb containing methane gas at a pressure of 2.96 atm, is opened and the gases are allowed to mix. Assuming that the temperature remains constant, the final pressure in the system is ?? atm.

Answer 1

There are a couple of ways to answer this question. It may be easiest to think of two separate systems at first, and to then combine them later. So we could find the moles of helium in one system and the moles of methane in the other and then combine them at the end. Let's do that using PV = nRT (ideal gas law) and we can solve for n (moles) and since temperature (T) is constant we can just make it 273K for convenience.

n = PV/RT

For Helium: n = (6.06 atm)(2.64 L)/(0.0821 Latm/Kmol)(273K) = 0.714 moles He

For Methane: n = (2.96 atm)(7.11 L)/(0.0821 Latm/Kmol)(273) = 0.939 moles CH4

Total moles - 0.714 + 0.939 = 1.653 moles

Total volume = 2.64 L + 7.11 L = 9.75 L

Solving the ideal gas law for final pressure, we have...

PV = nRT

P = nRT/V =(1.653 mol)(0.0821 Latm/Kmol)(273)/9.75 L

P = 3.80 atm

Or using Daltons law of partial pressures, we have...

Total final volume = 9.75 L

6.06 atm x 2.64 L /9.75 L = 1.64 atm for He

2.96 atm x 7.11 L /9.75 L = 2.16 atm for methane

Total pressure = 1.64 atm + 2.16 atm = 3.80 atm