Charles's Law

Submitted by ChemPRIME Staff on Thu, 12/16/2010 - 13:22
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  • While Boyle's Law explores the effect of pressureForce per unit area; in gases arising from the force exerted by collisions of gas molecules with the wall of the container. on the volume of a gas, Charles' Law examines the effect of temperatureA physical property that indicates whether one object can transfer thermal energy to another object.. Again you are probably familiar with the fact that increasing the temperature of a gas will cause the gas to expand. This effect was first studied quantitatively in 1787 by Jacques Charles (1746 to 1823) of France. Typical data from such an experiment are given in Table 1. You can see that for 0.0466 mol H2(g) at constant pressure, a 50°C rise in temperature produces a 0.18-literA unit of volume equal to a cubic decimeter. increase in volume, whether the temperature increases from 0.0 to 50.0° or from 100.0 to 150.0°C.


    TABLE 1 Variation in the Volume of H2(g) with Temperature.

    Temperature (in degrees C) Volume / L
    Data for 0.0446 mol H2(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 H2(g) 1 atm (101.3 kPa)
    0.0 2.24
    50.0 2.65
    100.0 3.06
    150.0 3.47


    When the experimental data of Table 1 are graphed, we obtain Fig. 1. Notice that the four points corresponding to 0.0446 mol H2(g) lie on a straight line, as do the points for 0.100 mol H2(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 H2(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 substanceA material that is either an element or that has a fixed ratio of elements in its chemical formula. would condense to a liquidA state of matter in which the atomic-scale particles remain close together but are able to change their positions so that the matter takes the shape of its container and freeze to a solidA state of matter having a specific shape and volume and in which the particles do not readily change their relative positions. as it was cooled. When the pressure is 1.00 atm (101.3 kPa), H2(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 zeroThe minimum possible temperature: 0 K, -273.15 °C, -459.67 °F. of temperature—it is impossible to go any lower.

    In Fig. 1b 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 unitsThe international system of units (Système International d'Unité) based on seven fundamental units: meter, kilogram, second, ampere, kelvin, candela, mole. called kelvins (abbreviated K), in honor of the English physicist William Thomson, Lord Kelvin (1824 to 1907).

    Figure 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 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 scaleis (0 K) is –273.15°C. The freezing pointThe temperature at which a liquid becomes a solid; also called melting 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. 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      (1)


    As in the case of previous gas laws, we can introduce a proportionality constant, in this case, kC:


    V=k_{\text{C}}T\text{        or          }\frac{V}{T}=k_{\text{C}}\text{                        (2)}



    EXAMPLE 1 A sample of H2(g) occupies a volume of 69.37 cm³ at a pressure of exactly 1 atm when immersed in a mixtureA combination of two or more substances in which the substances retain their chemical identity. of ice and water. When the gas (at the same pressure) is immersed in boilingThe process of a liquid becoming vapor in which bubbles of vapor form beneath the surface of the liquid; at the boiling temperature the vapor pressure of the liquid equals the pressure of the gas in contact with the liquid. benzene, its volume expands to 89.71 cm3. What is the boiling pointThe temperature at which the vapor pressure of a liquid equals the pressure of the gas in contact with the liquid; usually this is atmospheric pressure. of benzene?


    SolutionA mixture of one or more substances dissolved in a solvent to give a homogeneous mixture. 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 freezingThe process of forming a solid from a liquid. 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.



    Because both temperature and pressure affect the volume of a gas, it is convenient to specify a reference temperature and pressure at which all volumes may be compared. Standard temperature and pressure (or STP) is chosen to be 273.15 K (0°C) and 1 atm (101.325 kPa)