This chapter is from the book After Boyle’s Law was published by Robert Boyle in 1662, it was postulated by Jacques Charles in 1780 and confirmed in 1802 that the volume of a given mass of a gas would vary according to the absolute temperature of the gas. At this time, no one knew the true nature of gases. Charles’s Law states that under the condition of constant pressure the volume of a given mass of an ideal gas is directly proportional to its absolute temperature. Put into a mathematical form, Charles’s Law becomes Equation 4.6. Equation 4.6
 In this equation, V is the volume of a fixed amount of gas in terms of either moles or mass and T is the absolute temperature. The linear relationship between volume and temperature is seen in Figure 4.2.
Figure 4.2 Plot of volume V versus absolute temperature T for a gas at a constant pressure Just as we did before, Equation 4.6 can be rewritten and then used to show volumes before and after a change of temperature, which results in Equation 4.7. Equation 4.7
The subscripts 1 and 2 indicate the states before and after a change. Remember, absolute temperatures must be used. As mentioned in Chapter 3, “Units of Measure,” the absolute temperature scale is measured relative from absolute zero where all motion ceases. In the Celsius temperature scale, where the freezing point of pure water is defined as zero degrees Celsius (0°C) and the boiling point of water at 1 atmosphere of pressure is defined as one hundred degrees Celsius (100°C) absolute zero would be approximately –273.15°C. For the Fahrenheit scale, with 32°F and 212°F being defined as the freezing and boiling points, respectively, absolute zero is –459.67°F. Absolute temperature is discussed in more detail in the next section. As with Boyle’s Law, Equation 4.7 can be rewritten to give Equation 4.8. Equation 4.8
Alternatively, it can be arranged as shown in Equation 4.9. Equation 4.9
NOTE** Charles Law problems must have the PRESSURE CONSTANT. Try using Charles’s law to solve the following problem. Example A sample of gas at 15ºC and 1 atm has a volume of 2.50 L. What volume will this gas occupy at 30ºC and 1 atm? OK, Here is our formula, NO As a Chemistry teacher I am REQUIRED to give you the temperature in Celsius and make you convert to Kelvin. It is the Law of all Chemistry Teachers. So convert the temperatures to Kelvin. T1= 15C +273=288K and T2 =30C +273=303K Now Plug in and Solve
This makes sense—the temperature is increasing slightly, so the volume should increase slightly. Again be careful of questions like this. It’s tempting to just use the Celsius temperature, but you must first convert to Kelvin temperature (by adding 273) to get the correct relationships!
A fun laboratory demonstration in which a fully inflated balloon is placed in liquid nitrogen (at a temperature of –196 ˚C) and it shrinks to about 1/1000th of its former size. If the balloon is carefully removed and allowed to warm to room temperature, it will again be fully inflated. This is a simple demonstration of the effect of temperature on the volume of a gas. In 1787, Jacques Charles performed a systematic study of the effect of temperature on gases. Charles took samples of gases at various temperatures, but at the same pressure, and measured their volumes. The first thing to note is that the plot is linear. When the pressure is constant, volume is a direct linear function of temperature. This is stated as Charles’s law.
The volume (V) of an ideal gas varies directly with the temperature of the gas (T) when the pressure (P) and the number of moles (n) of the gas are constant. We can express this mathematically as: \[V\propto T\; \; at\; constant\; P\; and\; n \nonumber \] \[V=constant(T)\; or\; \frac{V}{T}=constant \nonumber \]V ∝ T ( at constant P and n ) {\displaystyle V\propto T{\text{ }}\left({\text{at constant }}P{\text{ and }}n\right)} The data for three different samples of the same gas is as follows: 0.25 moles, 0.50 moles and 0.75 moles. All of these samples behave as predicted by Charles’s law (the plots are all linear), but , if you extrapolate each of the lines back to the y-axis (the temperature axis), all three lines intersect at the same point! This point, with a temperature of –273.15 ˚C, is the theoretical point where the samples would have “zero volume”. This temperature, -273.15 ˚C, is called absolute zero. An even more intriguing thing is that the value of absolute zero is independent of the nature of the gas that is used. Hydrogen, oxygen, helium, argon, (or whatever), all gases show the same behavior and all intersect at the same point. The temperature of this intersection point is taken as “zero” on the Kelvin temperature scale. The abbreviation used in the Kelvin scale is K (no degree sign) and there are never negative values in degrees Kelvin. The size of the degree increment in Kelvin is identical to that in Centigrade and Kelvin and centigrade scales are related by the simple conversion: \[Kelvin=Centigrade+273.15 \nonumber \]
Please note that whenever you work gas law problems where temperature is one variable, you MUST use the Kelvin scale. Just like we did for pressure-volume problems, we can use Charles’s law to predict what will happen to the volume of a sample of gas as we change the temperature. Because V ╱ T {\displaystyle {}^{V}\!\!\diagup \!\!{}_{T}\;} is a constant for any given sample of gas (at constant P), we can again imagine two states; an initial state with a certain temperature and volume ( V 1 ╱ T 1 {\displaystyle {}^{V_{1}}\!\!\diagup \!\!{}_{T_{1}}\;} ), and a final state with different values for pressure and volume ([ V 2 ╱ T 2 {\displaystyle {}^{V_{2}}\!\!\diagup \!\!{}_{T_{2}}\;} \[\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}} \nonumber \]
We have a container with a piston that we can use to adjust the pressure on the gas inside. You are told that, initially, the temperature of the gas in the container is 175 K and the volume is 1.50 L. The temperature is changed to 76 K and the piston is then adjusted so that the pressure is identical to the pressure in the initial state; what is the final volume?
We substitute into our Charles’s law equation: \[\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}} \nonumber \] \[\frac{1.50\; L}{175\; K}=\frac{V_{2}}{76\; K} \nonumber \] \[V_{2}=\left ( \frac{(76\; K)(1.50\; L)}{175\; K} \right )=0.65\; L \nonumber \]
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What is the temperature at which volume of a gas will be zero at constant pressure according to Charles Law?
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