Chemistry Reference
In-Depth Information
We would convert the pressure to kPa by making use of the conversion
factor 1.00 kPa = 7.50 mm of Hg.
1.00 kPa
7.50 mm of Hg
888 mm of Hg ×
= 118 kPa
Now, substitute these values into the original problem and solve.
How much spac
e
would 3.45 moles of nitrogen gas occupy at a tem-
perature of 18
o
C 291 K and a pressure of 888 mm of Hg 118 kPa?
Given: n = 3.45 mol; T = 291 K; P = 118 kPa;
R = 8.31 dm
3
× kPa/mol × K
Find: V
Formula:
V
=
nRT
(3.45 mol)(8.31 dm
3
× kPa/mol × K)(291 K)
118 kPa
=
P
= 70.7 dm
3
As you can see, using the Ideal Gas Law is really as simple as isolating
the unknown. Don't be intimidated by the large number of units involved
in the calculations. Simply isolate the unknown, substitute, and solve. Try
the review problems, and check your answers at the end of the chapter.
Lesson 8-7 Review
In the Ideal Gas Law formula, shown here, identify what each letter
stands for.
PV = nRT
1. P_______________
4. T_______________
2. V_______________
5. R_______________
3. n_______________
Use the Ideal Gas Law to solve each of the following problems:
6.
At what temperature would 2.5 moles of argon gas occupy 60.0 dm
3
at a pressure of 98.4 kPa?
7.
How many moles of helium gas would occupy 33.5 dm
3
at 22.0
o
C and
1.50 atm?
8.
How much space would 6.0 moles of oxygen gas occupy at 112.0 kPa
of pressure and 33.0
o
C?
