Boyle's Law

Submitted by ChemPRIME Staff on Thu, 12/16/2010 - 13:21
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    • You are probably already familiar with the fact that when you squeeze a gasA state of matter in which a substance occupies the full volume of its container and changes shape to match the shape of the container. In a gas the distance between particles is much greater than the diameters of the particles themselves; hence the distances between particles can change as necessary so that the matter uniformly occupies its container., it will take up less space. In formal terms, increasing the pressureForce per unit area; in gases arising from the force exerted by collisions of gas molecules with the wall of the container. on a gas will decrease its volume. Studies in 1662 by the English scientist Robert Boyle (1627 to 1691) gave results such as those in Table 1. Careful, study of such data reveals that if we double the pressure, we halve the volume;


    TABLE 1 Variation in the Volume of 0.0446 mol H2(g) with Pressure at 0°C.

    Trial Pressure/kPa Pressure/atm Volume/liter
    1 152.0 1.50 0.666
    2 126.7 1.25 0.800
    3 101.3 1.00 1.00
    4 76.0 0.750 1.333
    5 50.7 0.500 2.00
    6 25.3 0.250 4.00
    7 10.1 0.100 10.00


    if we triple the pressure, the volume is reduced to one-third; and so on. In general, if we multiply the pressure by some factor x, then we divide the volume by the same factor x. Such a relationship, in which the increase in one quantity produces a proportional decrease in another, is called inverse proportionality.

    The results of Boyle’s experiments with gases are summarized in Boyle’s law—for a given amount of gas at constant temperatureA physical property that indicates whether one object can transfer thermal energy to another object., the volume is proportional to the pressure. In mathematical terms


    V\propto \frac{\text{1}}{P}\text{ (1)}


    The reciprocal of P indicates the inverse nature of the proportionality. Using the proportionality constant kA to convert relationship (1) to an equation, we have


    V=k_{\text{B}}\text{ }\times \text{ }\frac{\text{1}}{P}\text{=}\frac{k_{\text{B}}}{P}\text{ (2a)}

    Multiplying both sides of Eq.(2a) by P, we have


    PV = kB      (2b)


    where kA represents a constant value for any given temperature and amount (or massA measure of the force required to impart unit acceleration to an object; mass is proportional to chemical amount, which represents the quantity of matter in an object.) of gas.


    EXAMPLE 1 Using the data in red in Table 1, confirm that Boyle’s law is obeyed.

    SolutionA mixture of one or more substances dissolved in a solvent to give a homogeneous mixture. Since the data apply to the same amount of gas at the same temperature, PV should be constant [Eq. (2b)] if Boyle’s law holds.


    P1V1 = 1.50 × 0.666 liter = 0.999 atm liter

    P4V4 = 0.750 atm × 1.333 liter = 1.000 atm liter

    P6V6 = 0.250 atm × 4.00 liter = 1.00 atm liter


    The first productA substance produced by a chemical reaction. differs from the last two in the fourth significant digit. Since some data are reported only to three significant figures, PV is constant within the limits of the measurements.


    If the units atmosphereA unit of pressure equal to 101.325 kPa or 760 mmHg; abbreviated atm. Also, the mixture of gases surrounding the earth. liter, in which PV was expressed in Example 1, are changed to SI baseIn Arrhenius theory, a substance that increases the concentration of hydroxide ions in an aqueous solution. In Bronsted-Lowry theory, a hydrogen-ion (proton) acceptor. In Lewis theory, a species that donates a pair of electrons to form a covalent bond. units, an interesting result arises:


    \begin{align} & \text{1 atm }\times \text{ 1 liter}=\text{101}\text{.3 kPa }\times \text{ 1 dm}^{\text{3}} \ & \text{ }=\text{101}\text{.3 }\times \text{ 10}^{\text{3}}\text{ Pa }\times \text{ 1 dm}^{\text{3}}\text{ }\times \text{ }\left( \frac{\text{1 m}}{\text{10 dm}} \right)^{\text{3}} \ & \text{ }=\text{101}\text{.3 }\times \text{ 10}^{\text{3}}\text{ }\frac{\text{N}}{\text{m}^{\text{2}}}\text{ }\times \text{ 1 dm}^{\text{3}}\text{ }\times \text{ }\frac{\text{1 m}^{\text{3}}}{\text{10}^{\text{3}}\text{ dm}^{\text{3}}}\ & \text{ }=\text{101}\text{.3 N m}=\text{101}\text{.3 kg m s}^{-\text{2}}\text{ m}=\text{101}\text{.3 J} \ \end{align}


    In other words, PV has the same units (joules) as an energyA system's capacity to do work.. While this does not guarantee that PV is an energy (be careful about relying on cancellation of units unless you know that a relationship between quantities exists), it does suggest that we should explore the possibility, covered in the section on Kinetic Theory of Gases The above argument also shows that the product of the units kilopascals times cubic decimeters is the unit joules. In subsequent discussions you will find it handy to remember that


    \text{1 kPa }\times \text{ 1 dm}^{\text{3}}=\text{1 J and 1 kPa}=\frac{\text{1 J}}{\text{1 dm}^{\text{3}}}\text{ and 1 dm}^{\text{3}}=\frac{\text{1 J}}{\text{1 kPa}}


    Boyle’s law enables us to calculate the pressure or volume of a gas under one set of conditions, provided we know the pressure and volume under a previous set of circumstances.


    EXAMPLE 2 The volume of a gas is 0.657 liter under a pressure of 729.8 mmHg. What volume would the gas occupy at atmospheric pressure (760 mmHg)? Assume constant temperature and amount of gas.

    Solution Two methods of solution will be given.


    a) Since PV must be constant,

          P1V1 = kB = P2V2

    Initial conditions: P1 = 729.8 mmHgV1 = 0657 liter

    Final conditions: P2 = 760 mmHgV2 = ?


    Solving for V2, we have


    V_{\text{2}}=\frac{P_{\text{1}}V_{\text{1}}}{P_{\text{2}}}=\frac{\text{729}\text{.8 mmHg }\times \text{ 0}\text{.657 liter}}{\text{760 mmHg}}=\text{0}\text{.631 liter}


    b) Note that in method a the original volume was multiplied by a ratio of pressures (P1/P2):


    V2 = 0.657 liter × ratio of pressures


    Rather than solving algebraically, we can use common sense to decide which of the two possible ratios


    \frac{\text{729}\text{.8 mmHg}}{\text{760 mmHg}}\text{ or }\frac{\text{760 mmHg}}{\text{729}\text{.8 mmHg}}


    should be used. The units cancel in either case, and so units are no help. However, if you reread the problem, you will see that we are asked to find the new volume (V2) produced by an increase in pressure. Therefore there must be a decrease in volume, and we multiply the original volume by a ratio which is less than 1:


    V_{\text{2}}=\text{0}\text{.657 liter }\times \frac{\text{729}\text{.8 mmHg}}{\text{760 mmHg}}=\text{0}\text{.631 liter}


    It is reassuring that both common sense and algebra produce the same answer.