Untitled - Kelly Walsh High School

Gases 91
Note that the mandatory conversion to Kelvin is present. We will need to do a
pressure conversion since P 1 and P 2 do not have the same units. We can do the
pressure conversion at any time; however, we will temporarily postpone this
step. The presence of two values for any variable strongly indicates that we
need to use the combined gas law [(P 1V 1/T 1) (P 2V 2/T 2)]. In this case, we cannot
eliminate any of the variables. We will, therefore, go straight to the rearranging
step to isolate V 2. This rearrangement gives V 2 (P 1V 1T 2/T 1P 2). We
can now enter the values form the table into this equation. In addition, we will
add a step to cover the pressure conversion.
V 2 (P 1 )(V 1 )(T 2 )
(T 1 )(P 2 )
Quick Tip
Many students get into trouble because they place conversions randomly about
the page. If you do the conversion step as we did in this example, you will be
less likely to “lose” your conversion, and less likely to copy a value incorrectly.
Let’s try another problem: What volume does 2.05 mol of oxygen occupy at a
temperature of 25C and a pressure of 0.950 atm? As usual, we begin by extracting
the numbers (and associated units) and the desired unknown from the
remainder of the problem.
P 0.950 atm T 25C 298 K n 2.05 mol V ?
Since a second set of conditions is not present, we will most likely need to use
the ideal gas equation (PV nRT). We need to rearrange this equation to isolate
the unknown (V). Then we enter the appropriate values, including R, the
ideal gas constant:
V nRT
P
(745 torr) (2.50 L) (318 K)
(298 K) (0.750 atm)
1 atm
a b 3.4868 3.49 L
760 torr
L atm
(2.05 mol) a0.0821 b (298 K)
mol K
52.7946 52.8 L
(0.950 atm)