Hello, Kamren,
There are two possible approaches to answering this. I'll start with the harder one - it is the approach one needs when the gas is not at STP, as we have here.
1. The ideal gas law
PV = nRT, where P is pressure, V is volume, n is moles, R is the gas constant, and T is the temperature in Kelvin.
First let's determine how many moles of CH4 are in 97.2 grams of CH4.
Next, make a table of the values we need for the equation:
Finally, rearrange the ideal gas law to solve for V and then enter the values from the table. Check to see if the units cancle and onky volume (liters) is remaining. This gives us 136.2 liters. Seems like a lot for only 6 moles of gas, but gasses occupy a priviledged ststus among states of matter. In fact, one of their unique features can be found in how we can answer the problem a lot more simply, whcih is next:
2. Ideal Gas Volume at STP
One mole of any gas at STP occupies 22.4 Liters. Yes, it's true. This makes problems like this super easy to solve, If we have 6.075 moles of a gas at STP, we have [22.4(L/mole)*(6.075 moles)] = 136.2 Liters.
Oh . . . LOOK! This is the same value we got from the Ideal Gas Law calculation. Wow. Remember it: 22.4 l liters/mole of ANY gas at STP.
Very useful information when trying to impress your friends or teacher. It can also cut test times dramatically. .
Bob
