9.7: Charles's Law
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- 49443
TABLE \(\PageIndex{1}\) Variation in the Volume of H_{2}(g) with Temperature.
Temperature (degree C) | Volume (L) |
---|---|
Data for 0.0446 mol H_{2}(g) at 1 atm(101.3 kPa) | |
0.0 | 1.00 |
50.0 | 1.18 |
100.0 | 1.37 |
150.0 | 1.55 |
Data for 0.100 mol H_{2}(g) 1 atm (101.3 kPa) | |
0.0 | 2.24 |
50.0 | 2.65 |
100.0 | 3.06 |
150.0 | 3.47 |
Kelvin
When the experimental data of Table \(\PageIndex{1}\) are graphed, we obtain Fig. \(\PageIndex{1}\) . Notice that the four points corresponding to 0.0446 mol H_{2}(g) lie on a straight line, as do the points for 0.100 mol H_{2}(g). If the lines are extrapolated (extended beyond the experimental points) to very low temperatures, we find that both of them intersect the horizontal axis at –273°C. The behavior of H_{2}(g) (and of many other gases) at normal temperatures suggests that if we cool a gas sufficiently, its volume will become zero at –273°C.
Of course a real substance would condense to a liquid and freeze to a solid as it was cooled. When the pressure is 1.00 atm (101.3 kPa), H_{2}(g) liquefies at –253°C and freezes at –259°C, and so all experiments involving would have to be performed above –253°C. If we could find a gas that did not condense, however, it would still be impossible to cool it below –273°C, because at that temperature its volume would be zero. Going to a lower temperature would correspond to a negative volume—something that is very hard to conceive of. Hence –273°C is referred to as the absolute zero of temperature—it is impossible to go any lower.
In Fig. \(\PageIndex{1}\) b the zero of the temperature axis has been shifted to absolute zero. The temperature scale used in this graph is called absolute or thermodynamic temperature. It is measured in SI units called Kelvins (abbreviated K), in honor of the English physicist William Thomson, Lord Kelvin (1824 to 1907).
Figure \(\PageIndex{1}\) Charles' law. (a) Volume plotted against Celsius temperature for two samples of H(g) at 1.00 atm (101.3 kPa); (b) volume plotted against thermodynamic temperature (Kelvin) for the same two samples. Note how much simpler graph (b) is than graph (a).
The temperature interval 1 K corresponds to a change of 1°C, but zero on the thermodynamic scale is (0 K) is –273.15°C. The freezing point of water at 1.00 atm (101.3 kPa) pressure is thus 273.15 K. By shifting to the absolute temperature scale, we have simplified the graph of gas volume versus temperature.
Charles' Law
Figure 1b shows that the volume of a gas is directly proportional to its thermodynamic temperature, provided that the amount of gas and the pressure remain constant. This is known as Charles’ law, and can be expressed mathematically as where T represents the absolute temperature (usually measured in Kelvins).
\[V \propto T\]
As in the case of previous gas laws, we can introduce a proportionality constant, in this case, k_{C}:
\[V=k_{\text{C}}T\text{ or }\frac{V}{T}=k_{\text{C}}\text{ (2)}\]
As an additional resource, the Concord Consortium has a tool that allows you to change the temperature of a gas in a container and observe the resulting change in volume. This tool can help to cement your understanding of Charles' Law and the relation between volume and temperature: Charles' Law Interactive.
Example \(\PageIndex{1}\) : Boiling Point
A sample of H_{2}(g) occupies a volume of 69.37 cm³ at a pressure of exactly 1 atm when immersed in a mixture of ice and water. When the gas (at the same pressure) is immersed in boiling benzene, its volume expands to 89.71 cm^{3}. What is the boiling point of benzene?
Solution As in the case of Boyle’s law, two methods of solution are possible.
a) Algebraically, we have, from Eq. (2),
\(\frac{V_{\text{1}}}{T_{\text{1}}}=k_{\text{C}}=\frac{V_{\text{2}}}{T_{\text{2}}}\)
and substituting into the equation
\(T_{\text{2}}=\frac{V_{\text{2}}T_{\text{1}}}{V_{\text{1}}}=\frac{\text{89}\text{.71 cm}^{\text{3}}\text{ }\times \text{ 273}\text{.15 K}}{\text{69}\text{.37 cm}^{\text{3}}}=\text{353}\text{.2 K}\)
yields the desired result. (The ice-water mixture must be at 273.15 K, the freezing point of water.)
b) By common sense we argue that since the gas expanded, its temperature must have increased. Thus
\(T_{\text{2}}=\text{273}\text{.15 K }\times \text{ ration greater than 1}=\text{273}\text{.15 K }\times \text{ }\frac{\text{89}\text{.71 cm}^{\text{3}}}{\text{69}\text{.37 cm}^{\text{3}}}=\text{353}\text{.2 K}\)
Note: In this example we have used the expansion of a gas instead of the expansion of liquid mercury to measure temperature.
Contributors
Ed Vitz (Kutztown University), John W. Moore (UW-Madison), Justin Shorb (Hope College), Xavier Prat-Resina (University of Minnesota Rochester), Tim Wendorff, and Adam Hahn.
Kelvin When the experimental data of Table \(\PageIndex{1}\) are graphed, we obtain Fig. \(\PageIndex{1}\) . Notice that the four points corresponding to 0.0446 mol H_{2}(g) lie on a straight line, as do the points for 0.100 mol H_{2}(g). If the lines are extrapolated (extended beyond the experimental points) to very low temperatures, we find that both of them intersect the horizontal axis at –273°C. The behavior of H_{2}(g) (and of many other gases) at normal temperatures suggests that if we cool a gas sufficiently, its volume will become zero at –273°C.
Of course a real substance would condense to a liquid and freeze to a solid as it was cooled. When the pressure is 1.00 atm (101.3 kPa), H_{2}(g) liquefies at –253°C and freezes at –259°C, and so all experiments involving would have to be performed above –253°C. If we could find a gas that did not condense, however, it would still be impossible to cool it below –273°C, because at that temperature its volume would be zero. Going to a lower temperature would correspond to a negative volume—something that is very hard to conceive of. Hence –273°C is referred to as the absolute zero of temperature—it is impossible to go any lower.
In Fig. \(\PageIndex{1}\) b the zero of the temperature axis has been shifted to absolute zero. The temperature scale used in this graph is called absolute or thermodynamic temperature. It is measured in SI units called Kelvins (abbreviated K), in honor of the English physicist William Thomson, Lord Kelvin (1824 to 1907).
Figure \(\PageIndex{1}\) Charles' law. (a) Volume plotted against Celsius temperature for two samples of H(g) at 1.00 atm (101.3 kPa); (b) volume plotted against thermodynamic temperature (Kelvin) for the same two samples. Note how much simpler graph (b) is than graph (a).
The temperature interval 1 K corresponds to a change of 1°C, but zero on the thermodynamic scale is (0 K) is –273.15°C. The freezing point of water at 1.00 atm (101.3 kPa) pressure is thus 273.15 K. By shifting to the absolute temperature scale, we have simplified the graph of gas volume versus temperature.
Charles' Law
Figure 1b shows that the volume of a gas is directly proportional to its thermodynamic temperature, provided that the amount of gas and the pressure remain constant. This is known as Charles’ law, and can be expressed mathematically as where T represents the absolute temperature (usually measured in Kelvins).
\[V \propto T\] As in the case of previous gas laws, we can introduce a proportionality constant, in this case, k_{C}: \[V=k_{\text{C}}T\text{ or }\frac{V}{T}=k_{\text{C}}\text{ (2)}\]As an additional resource, the Concord Consortium has a tool that allows you to change the temperature of a gas in a container and observe the resulting change in volume. This tool can help to cement your understanding of Charles' Law and the relation between volume and temperature: Charles' Law Interactive.
Example \(\PageIndex{1}\) : Boiling Point
A sample of H_{2}(g) occupies a volume of 69.37 cm³ at a pressure of exactly 1 atm when immersed in a mixture of ice and water. When the gas (at the same pressure) is immersed in boiling benzene, its volume expands to 89.71 cm^{3}. What is the boiling point of benzene?
Solution As in the case of Boyle’s law, two methods of solution are possible.
a) Algebraically, we have, from Eq. (2), \(\frac{V_{\text{1}}}{T_{\text{1}}}=k_{\text{C}}=\frac{V_{\text{2}}}{T_{\text{2}}}\) and substituting into the equation \(T_{\text{2}}=\frac{V_{\text{2}}T_{\text{1}}}{V_{\text{1}}}=\frac{\text{89}\text{.71 cm}^{\text{3}}\text{ }\times \text{ 273}\text{.15 K}}{\text{69}\text{.37 cm}^{\text{3}}}=\text{353}\text{.2 K}\) yields the desired result. (The ice-water mixture must be at 273.15 K, the freezing point of water.) b) By common sense we argue that since the gas expanded, its temperature must have increased. Thus \(T_{\text{2}}=\text{273}\text{.15 K }\times \text{ ration greater than 1}=\text{273}\text{.15 K }\times \text{ }\frac{\text{89}\text{.71 cm}^{\text{3}}}{\text{69}\text{.37 cm}^{\text{3}}}=\text{353}\text{.2 K}\) Note: In this example we have used the expansion of a gas instead of the expansion of liquid mercury to measure temperature.
Contributors
Ed Vitz (Kutztown University), John W. Moore (UW-Madison), Justin Shorb (Hope College), Xavier Prat-Resina (University of Minnesota Rochester), Tim Wendorff, and Adam Hahn.