General Chemistry

1) Calculate the value of the gas law constant R when pressure is in atm, volume in Liter and temperature in
Kelvin.
Given R = 8.314 Jmol‐1K‐1, 1 atm = 1 .01325x105 Nm‐2, 1m3 = 1000 L

J

R = 8.314 mol x K

=

R = 8.314

2

m x mol x K

R = 0.082

atm x L mol x K

1.01325 x 10

2

m x mol x K

1

x

Nxm

=

mol x K

3

Nxm

3

Nxm

5

atm
Nm-2

x 1000

L
3

m

Reporting computed data
• Multiplication and Division
5.02 x 89.665 x 0.10 = 45.0118
How many significant figures should we take for the answer?

The relative uncertainty of each numbers =

45.0118 x

= 4.50

,

45.0118
(2 sig. figures)

,

Addition and subtraction
5.74 + 0.823 + 2.651 = 9.214
How many significant figures should we take for the answer?

5.74
+ 0.823
+ 2.651
9.214
9.21 (2 decimal places)

Logarithms and antilogarithms
• In a logarithm of a number, keep as many digits to the right of the decimal point as there are significant figures
• In antilogarithm of number, keep as many digits to the right of the decimal point in the original number.

• Anti‐log 12.5 = 3.16227 x 1012
1 decimal point
= 3 x 1012
(1 significant figure)

• Log 4.000 x 10‐5 = ‐4.3979400
(4 significant figures)
= ‐4.3979
4 decimal points

•

If either the unrounded result or the result rounded has 1 as its leading significant digit, and

.

the operand's leading significant digit isn't 1, keep an extra significant figure in the result

• Example A: (3.9)2 = 15.2
• Example B: 0.0144 = 0.120
• Example C: (4 x 101)2 = 1.6 x 103

Exercises
• The temperature in the stratosphere is ‐18oC. Calculate the root‐mean‐square speed of N2, O2 and O3 molecules in this region.(molecular mass of N2 = 28 g mol‐1, O2 = 32 g mol‐1, O3 species = 48 g mol‐1) rms =
For N2, the molar mass of N2 = 28 g mol‐1 rms =

.

.

.

J = kg m2s‐2
= 227150.3571
= 2.27x10 ms‐1
=4.7x102 ms‐1