Everything Science Grade 11

5.3 CHAPTER 5. THERMAL PROPERTIES AND IDEAL GASES

Step 2 : Convert the known values to SI units if necessary.

Here, temperature mustbe converted into Kelvin, therefore:

T 1 = 32 + 273 = 305 K

T 2 = 7 + 273 = 280 K

Step 3 : Choose a relevant gas law equation that will allow you to calculate theunknown

variable.

V 1

T 1

=

V 2

T 2

Therefore,

V 2 =

V 1 × T 2

T 1

Step 4 : Substitute the known values into the equation.Calculate the unknownvariable.

V 2 =

122 ml× 280 K

305 K

= 112 ml

Tip

Note that here the
temperature must be
converted to Kelvin
(SI) since the change
from degrees Celsius
involves addition, not
multiplication by a fixed
conversion ratio (as is
the case with pressure
and volume.)

Example 4: Charles’s Law 2

QUESTION

At a temperature of 298 K, a certain amountof CO 2 gas occupies a volumeof 6 �. What

volume will the gas occupy if its temperature isreduced to 273 K?

SOLUTION

Step 1 : Write down all the information that you knowabout the gas.

V 1 = 6 � and V 2 =?

T 1 = 298 K and T 2 = 273 K

Step 2 : Convert the known values to SI units if necessary.

Temperature data is already in Kelvin, and so noconversions are necessary.

Step 3 : Choose a relevant gas law equation that will allow you to calculate theunknown

variable.

V 1

T 1

=

V 2

T 2

Therefore,

V 2 =

V 1 × T 2

T 1

Step 4 : Substitute the known values into the equation.Calculate the unknownvariable.