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

The absolute uncertainty is then

= 227150.3571 x

= 12619

The result should be rounded to

227000 = 2.3 x 105

To avoid rounding errors, at least one extra digit should be kept through all the computations

Exercises

• The temperature in the stratosphere is ‐18.0oC. Calculate the root‐mean‐square speed of N2, O2 and O3 molecules in this region.(molecular mass of N2 = 28.0 g mol‐1, O2 = 32.0 g mol‐1, O3 species = 48.0 g mol‐1)

rms =

For N2, the molar mass of N2 = 28.0 g mol‐1

.

rms =

.

.

.

J = kg m2s‐2

= 227150.3571

= 2.271x10 ms‐1

= 477 ms‐1

The absolute uncertainty is then

= 227150.3571 x

= 1262

The result should be rounded to

227100 = 2.27 x 105

To avoid rounding errors, at least one extra digit should be kept through all the computations