Learning Objectives Define the relationship between gas temperature and volume, Charles's Law. Use Charles's Law to calculate changes in temperature or volume of a gas. In addition to pressure and volume, another measurable characteristics of a gas is temperature ( T ). Perhaps one can vary the temperature of a gas sample and note what effect it has on the other properties of the gas. Early scientists did just this, discovering that if the amount of a gas and its pressure are kept constant, then changing the temperature changes the volume ( V ). As temperature increases, volume increases; as temperature decreases, volume decreases. We say that these two characteristics are directly related . Kelvin Temperature Scale A mathematical relationship between V and T should be possible except for one thought: what temperature scale should we use? We know from previous chapters that scientists uses several possible temperature scales. Experiments show that the volume of a gas is related to its absolute temperature in Kelvin , not its temperature in degrees Celsius . If the temperature of a gas is expressed in Kelvins, then experiments show that the ratio of volume to temperature is a constant ( k ): \[\frac{V}{T}=k\nonumber \] Charles's Law We can modify this equation as we modified Boyle's law: the initial conditions V 1 and T 1 have a certain value, and the value must be the same when the conditions of the gas are changed to some new conditions V 2 and T 2 , as long as pressure and the amount of the gas remain constant. Thus, we have another gas law: \[\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}\; (at\; constant\; P\; and\; n)\nonumber \] This gas law is commonly referred to as Charles's law , after the French scientist Jacques Charles, who performed experiments on gases in the 1780s. The tactics for using this mathematical formula are similar to those for Boyle's law. To determine an unknown quantity, use algebra to isolate the unknown variable by itself and in the numerator; the units of similar variables must be the same. But we add one more tactic: all temperatures must be expressed in the absolute temperature scale (Kelvin). As a reminder, we review the conversion between the absolute temperature scale and the Celsius temperature scale: K = °C + 273 where K represents the temperature in kelvins, and °C represents the temperature in degrees Celsius. Example \(\PageIndex{1}\) A sample of gas has an initial volume of 34.8 mL and an initial temperature of 315 K. What is the new volume if the temperature is increased to 559 K? Assume constant pressure and amount for the gas. Solution First, we assign the given values to their variables. The initial volume is V 1 , so V 1 = 34.8 mL, and the initial temperature is T 1 , so T 1 = 315 K. The temperature is increased to 559 K, so the final temperature T 2 = 559 K. We note that the temperatures are already given in kelvins, so we do not need to convert the temperatures. Substituting into the expression for Charles's law yields \[\frac{34.8\, ml}{315\, K}=\frac{V_{2}}{559\, K}\nonumber \] We solve for V 2 by algebraically isolating the V 2 variable on one side of the equation. We do this by multiplying both sides of the equation by 559 K (number and unit). When we do this, the temperature unit cancels on the left side, while the entire 559 K cancels on the right side: \[\frac{(559\cancel{K})(34.8\, ml)}{315\, \cancel{K}}=\frac{V_{2}(\cancel{559\, K})}{\cancel{559\, K}}\nonumber \] The expression simplifies to \[\frac{(559)(34.8\, ml)}{315}=V_{2}\nonumber \] By multiplying and dividing the numbers, we see that the only remaining unit is mL, so our final answer is V 2 = 61.8 mL Does this answer make sense? We know that as temperature increases, volume increases. Here, the temperature is increasing from 315 K to 559 K, so the volume should also increase, which it does. Exercise \(\PageIndex{1}\) If V 1 = 3.77 L and T 1 = 255 K, what is V 2 if T 2 = 123 K? Answer 1.82 L It is more mathematically complicated if a final temperature must be calculated because the T variable is in the denominator of Charles's law. There are several mathematical ways to work this, but perhaps the simplest way is to take the reciprocal of Charles's law. That is, rather than write it as \[\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}\nonumber \] write the equation as \[\frac{T_{1}}{V_{1}}=\frac{T_{2}}{V_{2}}\nonumber \] It is still an equality and a correct form of Charles's law, but now the temperature variable is in the numerator, and the algebra required to predict a final temperature is simpler. Example \(\PageIndex{2}\) A sample of a gas has an initial volume of 34.8 L and an initial temperature of −67°C. What must the temperature of the gas be for its volume to be 25.0 L? Solution Here, we are looking for a final temperature, so we will use the reciprocal form of Charles's law. However, the initial temperature is given in degrees Celsius, not kelvins. We must convert the initial temperature to kelvins: −67°C + 273 = 206 K In using the gas law, we must use T 1 = 206 K as the temperature. Substituting into the reciprocal form of Charles's law, we get \[\frac{206\, K}{34.8\, L}=\frac{T_{2}}{25.0\, L}\nonumber \] Bringing the 25.0 L quantity over to the other side of the equation, we get \[\frac{(25.0\cancel{L})(206\, K)}{34.8\cancel{L}}=T_{2}\nonumber \] The L units cancel, so our final answer is T 2 = 148 K This is also equal to −125°C. As temperature decreases, volume decreases, which it does in this example. Exercise \(\PageIndex{2}\) If V 1 = 623 mL, T 1 = 255°C, and V 2 = 277 mL, what is T 2 ? Answer 235 K, or −38°C Summary Charles's law relates a gas's volume and temperature at constant pressure and amount. In gas laws, temperatures must always be expressed in kelvins.