Hello, Zoey,
We can use a gas law relationship in a case where there is no gain or loss of the number of moles of the gas. It is:
P1V1/T1 = P2V2/T2
Where P, V, and T are pressure, volume and temperature (in Kelvin), respectively. The subscripts 1 and 2 mean initial and final.
Make a table with the values and rearrange the equation to solve for the unknown, In this case, the unknown is P1.
P1 = P2(V2/V1)(T1/T2)
Please note that I’ve rearranged the equation to demonstrate that the initial pressure is the final pressure adjusted by the ratios of initial and final temperatures and pressures. (T1/T2) adjusts the volume to account for temperature differences.. (V2/V1) adjusts for volume differences.
Set up a table and enter the data. Pay attention to see if the units cancel to leave liters. Also be certain that temperatures are in Kelvin, which they are already, in this case.
Before doing the calculation, make a prediction as to what should happen. The volume grew a little (3.8 to 4.03 L) in spite of a large temperature decrease (320 to 236K). I'll approximate the two ratios as 1 and 1.4, to suggest:
P1 = (3.27)*(1)*(1.4), or around a 40% increase over 3.27.
This is too difficult to work out in my head, but I'd expect something in the range of 4.7 atm, or so. The point is that we'd expect to find that P1 must be greater than P2 to satisfy the temperature and volume differences.
Now we are ready to plug in the numbers from the table
P1 = P2(V2/V1)(T1/T2)
P1 = (3.27atm)*(4.03L/3.8L)*(320K/246K)
P1 = 4.702 atm
We guesstimated 4.7, so I'm comfortable I entered the correct numbers for the calculation.
Bob
