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Charles' Law Concepts
Consider the following experiment to measure the expansion of hydrogen gas:
 A known amount of hydrogen gas is drawn up into a syringe at 23^{o}C and 100 kPa pressure.
 The volume of hydrogen gas is recorded as 25 mL.
 The hydrogen gas in the syringe is then heated while maintaining a constant pressure of 100 kPa.
 The volume of hydrogen gas is recorded at various temperatures.
 The results of the experiment are shown in the table.
Volume and Temperature of a Quantity of H_{2(g)} at 100 kPa 

Volume (mL) 
Temperature (^{o}C) 
Trend 
25 
23 
As volume increases, temperature increases.
As volume decreases, temperature decreases.
As temperature increases, volume increases.
As temperature decreases, volume decreases.

30 
27 
35 
77 
40 
127.5 
45 
177 
Is there a simply relationship between the temperature of a gas in °C and its volume?
Let's try dividing volume by temperature and see, the results are in the table below:
Volume and Temperature of a Quantity of H_{2(g)} at 100 kPa 

Volume (mL) 
Temperature (^{o}C) 
V ÷ T(°C) 
25 
23 
1.1

30 
27 
1.1 
35 
77 
0.45 
40 
127.5 
0.31 
45 
177 
0.25 
No, there doesn't appear to be a simple relationship between the volume of the gas and its temperature in °C.
But, what happens if we convert all the temperatures in °C to temperatures in kelvin (K)?
The table below shows the temperature conversions:
Volume and Temperature of a Quantity of H_{2(g)} at 100 kPa 

Volume / mL 
Temperature / ^{o}C 
Temperature / K T(K) = 273 + T(°C) 
25 
23 
250 
30 
27 
300 
35 
77 
350 
40 
127.5 
400.5 
45 
177 
450 
Is there now a simple relationship between volume of gas and its temperature in Kelvin?
The table below shows the results of dividing the volume of a gas by its temperature in Kelvin:
Volume and Temperature of a Quantity of H_{2(g)} at 100 kPa 

Volume / mL 
Temperature / K T(K) = 273 + T(°C) 
V ÷ T (K) 
25 
250 
0.1 
30 
300 
0.1 
35 
350 
0.1 
40 
400.5 
0.1 
45 
450 
0.1 
Yes, we can now see a clear relationship between the volume of this gas (V) and its temperature in Kelvin (T) at a constant pressure of 100 kPa:
V ÷ T = 0.1
In general we could write:
V ÷ T = "a constant"
By rearranging this equation we can write:
V = "a constant" × T
Which is the equation for a straight line that goes through the origin (0,0) and has a slope (or gradient) equal to the value of "a constant".
The points are plotted and the line is extrapolated back to 0 (volume = 0 mL and temperature = 0 K) in the graph below:
volume (mL)

Expansion of Hydrogen Gas at Constant Pressure
temperature (K) 
The graph of gas volume against temperature is a straight line.
We say that there is a linear relationship between the volume of a gas and its temperature at constant pressure.
Extrapolation of the line back to (0,0) assumes that at temperatures below 23^{o}C (250 K), the linear relationship between volume and temperature will be maintained.
It is unlikely that this assumption will hold at very low temperatures for 100 kPa pressure as the hydrogen is likely to condense into a liquid first.
The extrapolation of the graph actually suggests that at 0 K an ideal gas has no volume (0 mL on our graph).
From the graph we see that:
Calculations : ^{Vi}/_{Ti} = ^{Vf}/_{Tf}
Consider an experiment in which we have a known quantity of gas in vessel such as a syringe in which the piston (plunger) can move freely up or down in order to change the volume occupied by the gas.
In the beginning of an experiment, a known amount of gas at a specified pressure has
volume = V_{i}
temperature = T_{i}
In the beginning of the experiment,
V_{i} ÷ T_{i} = constant = k
The temperature of the gaseous system is then changed (the system is heated or cooled) while constant pressure is maintained.
At the end of the experiment, the gas will have a different volume and a different temperature:
volume = V_{f}
temperature = T_{f}
As long as the amount of gas has not changed, and the pressure has not changed, then
V_{f} ÷ T_{f} = the same constant as at the beginning of the experiment = k
Therefore V_{i} ÷ T_{i} = k = V_{f} ÷ T_{f}
So V_{i} ÷ T_{i} = V_{f} ÷ T_{f}
This equation can then be rearranged to find the volume or temperature of a known amount of gas at specified pressure during the course of an experiment:
Find the initial volume, V_{i}
V_{i} = 
T_{i} × V_{f} T_{f} 
Find the final volume, V_{f}
V_{f} = 
V_{i} × T_{f} T_{i} 
Find the initial temperature, T_{i}
T_{i} = 
T_{f} × V_{i} V_{f} 
Find the final temperature, T_{f}
T_{f} = 
T_{i} × V_{f} V_{i} 
Worked Example of Charles' Law to Calculate Temperature of Gas
Question : A helium filled balloon had a volume of 75 L at 25^{o}C.
To what does the temperature, in Kelvin, need to be raised in order for the balloon to have a volume of 100 L at the same pressure?
Solution:
(Based on the StoPGoPS approach to problem solving.)
 What is the question asking you to do?
Calculate final temperature
T_{f} = ? K
 What data (information) have you been given in the question?
Extract the data from the question:
Conditions: constant amount of gas and pressure.
V_{i} = 75 L
T_{i} = 25°C
Convert temperature in °C to K
T_{i} + 273 = 25 + 273 = 298 K
V_{f} = 100 L
 What is the relationship between what you know and what you need to find out?
Assume ideal gas behaviour.
Because the amount of gas and pressure are constant, we can use Charles' Law:
V_{i} T_{i} 
= 
V_{f} T_{f} 
Multiply both sides of the equation by T_{f}:
T_{f} × V_{i} T_{i} 
= 
T_{f} × V_{f}
T_{f} 
T_{f} × V_{i} T_{i} 
= 
V_{f} 
Multiply both sides of the equation by T_{i}:
T_{i} × T_{f} × V_{i}
T_{i} 
= 
T_{i} × V_{f} 
T_{f} × V_{i} 
= 
T_{i} × V_{f} 
Divide both sides of the equation by V_{i}
T_{f} × V_{i}
V_{i} 
= 
T_{i} × V_{f} V_{i} 
T_{f} 
= 
T_{i} × V_{f} V_{i} 
 Substitute in the values and solve for T_{f}
T_{f} 
= 
T_{i} × V_{f} V_{i} 
T_{f} 
= 
298 × 100 75 

= 
397 K 
 Is your answer plausible?
Consider: V ∝ T (in K)
Increasing the volume of gas increases its temperature (in order to maintain constant pressure).
Double the the volume and the temperature (in K) doubles.
In the question the volume has increased by a factor of 100/75 = 1.333
So the temperature must have increased by a factor of 1.333
That is, final temperature = 1.333 × 298 = 397 K
Since this is the same value as we calculated above, we are reasonably confident that our answer is plausible.
 State your solution to the problem "calculate final temperature in K ":
T_{f} = 397 K