Everybody enjoys the smell and taste of freshly-baked bread. It is light and fluffy as a result of the action of yeast on sugar. The yeast converts the sugar to carbon dioxide, which at high temperatures causes the dough to expand. The end result is an enjoyable treat, especially when covered with melted butter. French physicist Jacques Charles (1746-1823) studied the effect of temperature on the volume of a gas at constant pressure. Charles's Law states that the volume of a given mass of gas varies directly with the absolute temperature of the gas when pressure is kept constant. The absolute temperature is temperature measured with the Kelvin scale. The Kelvin scale must be used because zero on the Kelvin scale corresponds to a complete stop of molecular motion. Mathematically, the direct relationship of Charles's Law can be represented by the following equation: \[\dfrac{V}{T} = k \nonumber \] As with Boyle's Law, \(k\) is constant only for a given gas sample. The table below shows temperature and volume data for a set amount of gas at a constant pressure. The third column is the constant for this particular data set and is always equal to the volume divided by the Kelvin temperature. When this data is graphed, the result is a straight line, indicative of a direct relationship, shown in the figure below. Notice that the line goes exactly toward the origin, meaning that as the absolute temperature of the gas approaches zero, its volume approaches zero. However, when a gas is brought to extremely cold temperatures, its molecules would eventually condense into the liquid state before reaching absolute zero. The temperature at which this change into the liquid state occurs varies for different gases. Charles's Law can also be used to compare changing conditions for a gas. Now we use \(V_1\) and \(T_1\) to stand for the initial volume and temperature of a gas, while \(V_2\) and \(T_2\) stand for the final volume and temperature. The mathematical relationship of Charles's Law becomes: \[\dfrac{V_1}{T_1} = \dfrac{V_2}{T_2} \nonumber \] This equation can be used to calculate any one of the four quantities if the other three are known. The direct relationship will only hold if the temperatures are expressed in Kelvin. Temperatures in Celsius will not work. Recall the relationship that \(\text{K} = \: ^\text{o} \text{C} + 273\). A balloon is filled to a volume of \(2.20 \: \text{L}\) at a temperature of \(22^\text{o} \text{C}\). The balloon is then heated to a temperature of \(71^\text{o} \text{C}\). Find the new volume of the balloon. Given: \(V_1 = 2.20 \: \text{L}\) and \(T_1 = 22^\text{o} \text{C} = 295 \: \text{K}\) \(T_2 = 71^\text{o} \text{C} = 344 \: \text{K}\) Find: V2 = ? L First, rearrange the equation algebraically to solve for \(V_2\). \[V_2 = \dfrac{V_1 \times T_2}{T_1} \nonumber \] Now substitute the known quantities into the equation and solve. \[V_2 = \dfrac{2.20 \: \text{L} \times 344 \: \cancel{\text{K}}}{295 \: \cancel{\text{K}}} = 2.57 \: \text{L} \nonumber \]
If V1 = 3.77 L and T1 = 255 K, what is V2 if T2 = 123 K? Answer1.82 L
A sample of a gas has an initial volume of 34.8 L and an initial temperature of −67°C. What must be the temperature of the gas for its volume to be 25.0 L?
If V1 = 623 mL, T1 = 255°C, and V2 = 277 mL, what is T2? Answer235 K, or −38°C Summary
LICENSED UNDER V ∝ T (in Kelvin) So at constant pressure, if the temperature (in Kelvin) is doubled, the volume of gas is also doubled. V ÷ T(K) = a constant or V/T(K) = a constant Vi ÷ Ti (K) = Vf ÷ Tf (K) or Vi/Ti = Vf/Tf Vi and Vf must be in the same units of measurement (eg, both in litres) A Real Gas is one which approaches Charles' Law as the temperature is raised or the pressure lowered. Please do not block ads on this website. Consider the following experiment to measure the expansion of hydrogen gas: As volume decreases, temperature decreases. As temperature increases, volume increases. As temperature decreases, volume decreases. Is there a simply relationship between the temperature of a gas in °C and its volume?
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)?
Is there now a simple relationship between volume of gas and its temperature in Kelvin?
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:
The graph of gas volume against temperature is a straight line. Extrapolation of the line back to (0,0) assumes that at temperatures below -23oC (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:
Do you know this? Join AUS-e-TUTE! Play the game now! 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 = Vi In the beginning of the experiment, Vi ÷ Ti = 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 = Vf As long as the amount of gas has not changed, and the pressure has not changed, then Vf ÷ Tf = the same constant as at the beginning of the experiment = k Therefore Vi ÷ Ti = k = Vf ÷ Tf So Vi ÷ Ti = Vf ÷ Tf 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, Vi Find the final volume, Vf Find the initial temperature, Ti Find the final temperature, Tf
Do you understand this? Join AUS-e-TUTE! Take the test now! Question : A sample of unknown gas had a volume of 1.2 L at 100oC and 100 kPa pressure. Solution: (Based on the StoPGoPS approach to problem solving.)
Can you apply this? Join AUS-e-TUTE! Do the drill now! Question : A helium filled balloon had a volume of 75 L at 25oC. Solution: (Based on the StoPGoPS approach to problem solving.)
Can you apply this? Join AUS-e-TUTE! Take the exam now! 1. Deviations from Ideal Gas Behaviour: As a Real Gas is cooled at constant pressure from a point well above its condensation point, its volume begins to increase linearly. As the temperature approaches the gases condensation point, the line begins to curve (usually downward) so there is a marked deviation from Ideal Gas behaviour close to the condensation point. Once the gas condenses to a liquid it is no longer a gas and so does not obey Charles' Law at all. |