Hi everybody, i'm having a little trouble with this stp and gas volume

ok heres what we should be aware with:

**as Temp increases, volume increases, as pressure increases vol decreases**

from kaplan book, they give an example of what seems to contradict this fact:

What is the molar mass of a 2L sample of gas that weighs 8g at a temp of 15C and a pressure of 1.5atm?

They do: V = (2L)(273/288)(1.5/1)

What they are doing is comparing the volume of the gas with the volume at STP (cool I get that), but why are they putting an increase in temp in the denominator? (as temp increases, volume increases is a fact) same for pressure (its reversed!) any insight would be great!

thnx!!

What they did here is find the volume of the gas at STP and the conditions the original gas was in.

You can find the volume of the gas by doing an initial/final state problem:

PV=nRT (STP)

--------

PV=nRT (original)

(1)(V)=nR(273)

--------------

(1.5)(2)=nR(288)

n and R cancel out.

It becomes V/3 = 273/288

Solve for V.

V (at STP)= 2.84375 L

That's how they got that V. Now if you keep going..8g/ 2.84375 L =

2.813 g/L: the density of the gas at STP.

Plus since we know that there are 22.4 mol/L of any gas at STP, multiply 22.4 by 2.813 so you can get rid of the L values. That step is key to understanding. You multiply them to get rid of L, and your answer is in terms of g/mol (the units of molar mass). So, 22.4 x 2.813 = 63.015 g/mol

So yea, there are a few ways to do this problem. Absolutely no reason to do more work than you have to do though...(meaning you don't even need to worry about STP at all)....just apply the ideal gas law. That way is faster and easier to comprehend.