Given: Ar P1: 3.5107 atm P2: 4.7902 atm V1: 63.805 L T1: 287.32 K T2: 512.3 K Wanted: liters of Ar?

Answer 1

Hello, Brandy,

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 P2.

V2 = V1(T2/T1)(P1/P2)

Please note that I’ve rearranged the equation to demonstrate that the final volume is simply the initial volume adjusted by the ratio of initial and final temperatures and pressures. (T2/T1) tells us that as temperature is increased, volume increases. (P1/P2) tells us that as pressure increases, volume decreases. Makes perfect sense, when expressed in this manner.

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 we do the calculation, make a prediction as to what should happen. Both temperature and pressure increase, in a closed chamber.  It will be hard to predict whether volume should increase or decrease, since the increases work against each another. Higher temperatures should increase the volume, while higher pressures will decrease it. But the ratio of pressures (P1/P2 = 3.5/4.8) will have less of an impact than the ratio of temperatures (T2/T1 = 512/287). So we should predict that the volume, V2, should increase less than double that of V1, judging by these two factors. I’ll take a wild guess of around 90 atm, just to set an expectation.

Now we are ready to plug in the numbers from the table

V2 = V1(T2/T1)(P1/P2)

V2 = (63.805L)*(512.3K/287.32K)*(3.5107atm/4.7902atm)

V2 = 83.38L

We guesstimated 90, so I'm satisfied.

Bob