Equations of state

Fundamentals

F.7 Equations of state

We have already remarked that the state of any sample of substance can be speci- fied by giving the values of the following properties:

V, the volume the sample occupies p, the pressure of the sample T, the temperature of the sample n, the amount of substance in the sample

However, an astonishing experimental fact is that these four quantities are not inde- pendent of one another. For instance, we cannot arbitrarily choose to have a sample of 0.555 mol H₂O in a volume of 100 cm³at 100 kPa and 500 K: it is found exper- imentallythat that state simply does not exist. If we select the amount, the volume, and the temperature, then we find that we have to accept a particular pressure (in this case, close to 230 kPa). The same is true of all substances, but the pressure in general will be different for each one. This experimental generalization is summa- rized by saying the substance obeys an equation of state, an equation of the form

pf(n,V,T) (F.8)

This expression tells us that the pressure is some function of amount, volume, and temperature and that if we know those three variables, then the pressure can have only one value.

The equations of state of most substances are not known, so in general we can- not write down an explicit expression for the pressure in terms of the other vari- ables. However, certain equations of state are known. In particular, the equation of state of a low-pressure gas is known and proves to be very simple and very use- ful. This equation is used to describe the behavior of gases taking part in reactions, the behavior of the atmosphere, as a starting point for problems in chemical engi- neering, and even in the description of the structures of stars.

We now pay some attention to gases because they are the simplest form of mat- ter and give insight, in a reasonably uncomplicated way, into the time scale of events on a molecular scale. They are also the foundation of the equations of ther- modynamics that we start to describe in Chapter 1, and much of the discussion of energy conversion in biological systems calls on the properties of gases.

The equation of state of a low-pressure gas was among the first results to be established in physical chemistry. The original experiments were carried out by COMMENT F.3 As

reviewed in Appendix 4, chemical amounts, n, are expressed in moles of specified entities.Avogadro’s constant, NA6.022 141 99

10²³mol¹, is the number of particles (of any kind) per mole of substance. ■

Low temperature

(a)

(b)

Energy as heat Equal temperature

High temperature

Fig. F.5 The temperatures of two objects act as a signpost showing the direction in which energy will flow as heat through a thermally conducting wall: (a) heat always flows from high temperature to low temperature. (b) When the two objects have the same temperature, although there is still energy transfer in both directions, there is no net flow of energy.

Robert Boyle in the seventeenth century, and there was a resurgence in interest later in the century when people began to fly in balloons. This technological progress demanded more knowledge about the response of gases to changes of pres- sure and temperature and, like technological advances in other fields today, that interest stimulated a lot of experiments.

The experiments of Boyle and his successors led to the formulation of the fol- lowingperfect gas equation of state:

pVnRT (F.9)

In this equation (which has the form of eqn F.8 when we rearrange it into pnRT/V), the gas constant, R, is an experimentally determined quantity that turns out to have the same value for all gases. It may be determined by evaluating RpV/nRTas the pressure is allowed to approach zero or by measuring the speed of sound (which depends on R). Values of Rin different units are given in Table F.2.

In SI units the gas constant has the value R8.314 47 J K¹mol¹

The perfect gas equation of state—more briefly, the “perfect gas law”—is so called because it is an idealization of the equations of state that gases actually obey.

Specifically, it is found that all gases obey the equation ever more closely as the pressure is reduced toward zero. That is, eqn F.9 is an example of a limiting law, a law that becomes increasingly valid as the pressure is reduced and is obeyed exactly at the limit of zero pressure.

A hypothetical substance that obeys eqn F.9 at allpressures is called a perfect gas.²From what has just been said, an actual gas, which is termed a real gas, be- haves more and more like a perfect gas as its pressure is reduced toward zero. In practice, normal atmospheric pressure at sea level (p

⬇

100 kPa) is already low enough for most real gases to behave almost perfectly, and unless stated otherwise, we shall always assume in this text that the gases we encounter behave like a per- fect gas. The reason why a real gas behaves differently from a perfect gas can be traced to the attractions and repulsions that exist between actual molecules and that are absent in a perfect gas (Chapter 11).

EXAMPLE F.2 Using the perfect gas law

A biochemist is investigating the conversion of atmospheric nitrogen to usable form by the bacteria that inhabit the root systems of certain legumes and needs

Table F.2

The gas constant in various units R 8.314 47 J K¹mol¹

8.314 47 L kPa K¹mol¹ 8.205 74 10² L atm K¹mol¹

62.364 L Torr K¹mol¹

1.987 21 cal K¹mol¹

2The term “ideal gas” is also widely used.

to know the pressure in kilopascals exerted by 1.25 g of nitrogen gas in a flask of volume 250 mL at 20°C.

Strategy For this calculation we need to arrange eqn F.9 (pVnRT) into a form that gives the unknown (the pressure, p) in terms of the information supplied:

p nR V T

To use this expression, we need to know the amount of molecules (in moles) in the sample, which we can obtain from the mass, m, and the molar mass, M, the mass per mole of substance, by using nm/M. Then, we need to convert the temperature to the Kelvin scale (by adding 273.15 to the Celsius temperature).

Select the value of Rfrom Table F.2 using the units that match the data and the information required (pressure in kilopascals and volume in liters).

Solution The amount of N₂molecules (of molar mass 28.02 g mol¹) present is

n_N2 mol

The temperature of the sample is T/K20273.15

Therefore, from pnRT/V,

n R T293 K

p

V250 mL

p435 kPa

Note how all units (except kPa in this instance) cancel like ordinary numbers.

A note on good practice:It is best to postpone the actual numerical calculation to the last possible stage and carry it out in a single step. This procedure avoids rounding errors.

SELF-TEST F.3 Calculate the pressure exerted by 1.22 g of carbon dioxide confined to a flask of volume 500 mL at 37°C.

Answer:143 kPa ■

It will be useful time and again to express properties as molar quantities, cal- culated by dividing the value of an extensive property by the amount of molecules.

An example is the molar volume,V_m, the volume a substance occupies per mole (1.25/28.02) mol (8.314 47 kPa L K¹mol¹) (20273.15 K)

0.250 L

1.25 28.02 1.25 g

28.02 g mol¹ m

MN₂

⎫ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎭ ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ ⎫ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎭

⎧ ⎪ ⎨ ⎪ ⎩

of molecules. It is calculated by dividing the volume of the sample by the amount of molecules it contains:

Volume of sample

Vm (F.10)

Amount of molecules (mol)

We can use the perfect gas law to calculate the molar volume of a perfect gas at any temperature and pressure. When we combine eqns F.9 and F.10, we get

VnRT/p

V_m (F.11)

This expression lets us calculate the molar volume of any gas (provided it is be- having perfectly) from its pressure and its temperature. It also shows that, for a given temperature and pressure, provided they are behaving perfectly, all gases have the same molar volume.

Chemists have found it convenient to report much of their data at a particu- lar set of standard conditions. By standard ambient temperature and pressure (SATP) they mean a temperature of 25°C (more precisely, 298.15 K) and a pres- sure of exactly 1 bar (100 kPa). The standard pressureis denoted p^両, so p^両1 bar exactly. The molar volume of a perfect gas at SATP is 24.79 L mol¹, as can be verified by substituting the values of the temperature and pressure into eqn F.11.

This value implies that at SATP, 1 mol of perfect gas molecules occupies about 25 L (a cube of about 30 cm on a side). An earlier set of standard conditions, which is still encountered, is standard temperature and pressure(STP), namely 0°C and 1 atm. The molar volume of a perfect gas at STP is 22.41 L mol¹.

We can obtain insight into the molecular origins of pressure and temperature, and indeed of the perfect gas law, by using the simple but powerful kinetic model of gases(also called the “kinetic molecular theory,” KMT, of gases), which is based on three assumptions:

1. A gas consists of molecules in ceaseless random motion (Fig. F.6).

2. The size of the molecules is negligible in the sense that their diameters are much smaller than the average distance traveled between collisions.

3. The molecules do not interact, except during collisions.

The assumption that the molecules do not interact unless they are in contact im- plies that the potential energy of the molecules (their energy due to their position) is independent of their separation and may be set equal to zero. The total energy of a sample of gas is therefore the sum of the kinetic energies (the energy due to motion) of all the molecules present in it. It follows that the faster the molecules travel (and hence the greater their kinetic energy), the greater the total energy of the gas.

The kinetic model accounts for the steady pressure exerted by a gas in terms of the collisions the molecules make with the walls of the container. Each colli- sion gives rise to a brief force on the wall, but as billions of collisions take place

RTp nRTnp Vn

Vn

Fig. F.6 The model used for discussing the molecular basis of the physical properties of a perfect gas. The pointlike molecules move randomly with a wide range of speeds and in random directions, both of which change when they collide with the walls or with other molecules.

every second, the walls experience a virtually constant force, and hence the gas ex- erts a steady pressure. On the basis of this model, the pressure exerted by a gas of molar mass Min a volume Vis

p (F.12)

wherecis the root-mean-square speed(r.m.s. speed) of the molecules and is de- fined as the square root of the mean value of the squares of the speeds, v, of the molecules. That is, for a sample consisting of Nmolecules with speeds v1,v₂,...,v_N, we square each speed, add the squares together, divide by the total number of mol- ecules (to get the mean, denoted by

具

...

典

), and finally take the square root of the result:

c

具

v²

典

^1/2

冢 冣

^{1/2 (F.13)}

DERIVATION F.2 The pressure according to the kinetic molecular theory

Consider the arrangement in Fig. F.7. When a particle of mass mthat is travel- ing with a component of velocity v_xparallel to the x-axis (v_x0 correspond- ing to motion to the right and v_x0 to motion to the left) collides with the wall on the right and is reflected, its linear momentum changes from m

兩

v_x

兩

be- fore the collision to m

兩

v_x

兩

after the collision (when it is traveling in the opposite direction at the same speed). The x-component of the momentum therefore changes by 2m

兩

v_x

兩

on each collision (the y-andz-components are un- changed). Many molecules collide with the wall in an interval t, and the to- tal change of momentum is the product of the change in momentum of each molecule multiplied by the number of molecules that reach the wall during the interval.

Next, we need to calculate that number. Because a molecule with velocity componentv_xcan travel a distance

兩

v_x

兩

talong the x-axis in an interval t, all the molecules within a distance

兩

v_x

兩

tof the wall will strike it if they are trav- eling toward it. It follows that if the wall has area A, then all the particles in a volumeA

兩

v_x

兩

twill reach the wall (if they are traveling toward it). The num- ber density, the number of particles divided by the total volume, is nN_A/V(where nis the total amount of molecules in the container of volume V and N_A is Avogadro’s constant), so the number of molecules in the volume A

兩

v_x

兩

t is (nN_A/V) A

兩

v_x

兩

t. At any instant, half the particles are moving to the right and half are moving to the left. Therefore, the average number of collisions with the wall during the interval tis¹⁄2nN_AA

兩

v_x

兩

t/V.

Newton’s second law of motion states that the force acting on a particle is equal to the rate of change of the momentum, the change of momentum divided by the interval during which it occurs. In this case, the total momentum change in the interval tis the product of the number we have just calculated and the change 2m

兩

v_x

兩

:

Momentum change 2m

兩

v_x

兩

nMAv_x²t

V nmAN_Av_x²t

V nN_AA

兩

v_x

兩

t

2V v₁²v₂² v_N² N nMc²

3V

Volume =l^vxlΔtA

l^vxlΔt

AreaA Won't

Will

x

Fig. F.7 The model used for calculating the pressure of a perfect gas according to the kinetic molecular theory. Here, for clarity, we show only the x-component of the velocity (the other two components are not changed when the molecule collides with the wall). All molecules within the shaded area will reach the wall in an intervaltprovided they are moving toward it.

COMMENT F.4 The velocity,v, is a vector, a quantity with both magnitude and direction. The magnitude of the velocity vector is the speed,v, given by v(vx2 vy2vz2)^1/2, where vx,vy, and vz, are the components of the vector along the x-,y-, and z-axes, respectively (see the illustration). The magnitude of each component, its value without a sign, is denoted

兩

...

兩

. For example,

兩

vx

兩

means the magnitude of vx. The linear momentum,p, of a particle of massmis the vector pmv with magnitude pmv.■

v_z

v_x v_y

v

whereMmN_A. Next, to find the force, we calculate the rate of change of momentum:

Force

It follows that the pressure, the force divided by the area, is Pressure

Not all the molecules travel with the same velocity, so the detected pressure, p, is the average (denoted

具

...

典

) of the quantity just calculated:

p

To write an expression of the pressure in terms of the root-mean-square speed, c, we begin by writing the speed of a single molecule, v, as v²v_x²v_y²v_z². Because the root-mean-square speed, c, is defined as c

具

v²

典

^1/2 (eqn F.13), it follows that

c²

具

v²

典 具

v_x²

典 具

v_y²

典 具

v_z²

典

However, because the molecules are moving randomly and there is no net flow in a particular direction, the average speed along xis the same as that in the y andzdirections. It follows that c²3

具

v_x²

典

. Equation F.12 follows when

具

v_x²

典

1⁄3c²is substituted into pnM

具

v_x²

典

/V.

The r.m.s. speed might at first encounter seem to be a rather peculiar measure of the mean speeds of the molecules, but its significance becomes clear when we make use of the fact that the kinetic energy of a molecule of mass mtraveling at a speedvisE_K¹⁄2mv², which implies that the mean kinetic energy,

具

E_K

典

, is the av- erage of this quantity, or ¹⁄2mc². It follows that

c

冢 冣

^{1/2 (F.14)}

Therefore, wherever cappears, we can think of it as a measure of the mean kinetic energy of the molecules of the gas. The r.m.s. speed is quite close in value to an- other and more readily visualized measure of molecular speed, the mean speed,c

苶

, of the molecules:

c

苶

(F.15)

For samples consisting of large numbers of molecules, the mean speed is slightly smaller than the r.m.s. speed. The precise relation is

c

苶 冢

³⁸

冣

^1/2c

^⬇

^{0.921c (F.16)}

v1v2 vN

N 2

具

E_K

典

m nM

具

vx2

典

V

nMv_x² V

nMAv_x² V Change of momentum

Time interval

For elementary purposes and for qualitative arguments, we do not need to distin- guish between the two measures of average speed, but for precise work the distinc- tion is important.

SELF-TEST F.4 Cars pass a point traveling at 45.00 (5), 47.00 (7), 50.00 (9), 53.00 (4), 57.00 (1) km h¹, where the number of cars is given in parentheses.

Calculate (a) the r.m.s speed and (b) the mean speed of the cars. (Hint:Use the definitions directly; the relation in eqn F.16 is unreliable for such small samples.) Answer: (a) 49.06 km h¹, (b) 48.96 km h¹

Equation F.12 already resembles the perfect gas equation of state, for we can rearrange it into

pV¹⁄3nMc² (F.17)

and compare it to pVnRT. Equating the expression on the right of eqn F.17 to nRTgives

1⁄3nMc²nRT

Then’s now cancel. The great usefulness of this expression is that we can rearrange it into a formula for the r.m.s. speed of the gas molecules at any temperature:

c

冢 冣

^{1/2 (F.18)}

Substitution of the molar mass of O₂ (32.0 g mol¹) and a temperature corre- sponding to 25°C (that is, 298 K) gives an r.m.s. speed for these molecules of 482 m s¹. The same calculation for nitrogen molecules gives 515 m s¹.

The important conclusion to draw from eqn F.18 is that the r.m.s. speed of mol- ecules in a gas is proportional to the square root of the temperature. Because the mean speed is proportional to the r.m.s. speed, the same is true of the mean speed. There- fore, doubling the temperature (on the Kelvin scale) increases the mean and the r.m.s. speed of molecules by a factor of 2^1/21.414… .

ILLUSTRATION F.2 The effect of temperature on mean speeds

Cooling a sample of air from 25°C (298 K) to 0°C (273 K) reduces the original r.m.s. speed of the molecules by a factor of

冢 冣

^1/2

冢 冣

^1/20.957

So, on a cold day, the average speed of air molecules (which is changed by the same factor) is about 4% less than on a warm day. ■

So far, we have dealt only with the average speed of molecules in a gas. Not all molecules, however, travel at the same speed: some move more slowly than the

273 298 273 K

298 K 3RT

M

average (until they collide and get accelerated to a high speed, like the impact of a bat on a ball), and others may briefly move at much higher speeds than the av- erage but be brought to a sudden stop when they collide. There is a ceaseless re- distribution of speeds among molecules as they undergo collisions. Each mole- cule collides once every nanosecond (1 ns10⁹s) or so in a gas under normal conditions.

The mathematical expression that tells us the fraction of molecules that have a particular speed at any instant is called the distribution of molecular speeds.

Thus, the distribution might tell us that at 20°C, 19 out of 1000 O₂molecules have a speed in the range between 300 and 310 m s¹, that 21 out of 1000 have a speed in the range 400 to 410 m s¹, and so on. The precise form of the distribution was worked out by James Clerk Maxwell toward the end of the nineteenth century, and his expression is known as the Maxwell distribution of speeds. According to Maxwell, the fraction fof molecules that have a speed in a narrow range between sand s s (for example, between 300 m s¹and 310 m s¹, corresponding to s300 m s¹ands10 m s¹) is

fF(s)s with F(s)4

冢 冣

^{3/2s2eMs2/2RT (F.19)}

This formula was used to calculate the numbers quoted above.

Although eqn F.19 looks complicated, its features can be picked out quite read- ily. One of the skills to develop in physical chemistry is the ability to interpret the message carried by equations. Equations convey information, and it is far more im- portant to be able to read that information than simply to remember the equation.

Let’s read the information in eqn F.19 piece by piece.

Before we begin, and in preparation for the large number of occurrences of ex- ponential functions throughout the text, it will be useful to know the shape of ex- ponential functions. Here we deal with two types, e^axand e^ax2. An exponential function of the form e^axstarts off at 1 when x0 and decays toward zero, which it reaches as xapproaches infinity (Fig. F.8). This function approaches zero more rapidly as aincreases. The function e^ax2is called a Gaussian function. It also starts off at 1 when x0 and decays to zero as xincreases, however, its decay is initially slower but then plunges down more rapidly than e^ax. The illustration also shows the behavior of the two functions for negative values of x. The exponential func- tion e^axrises rapidly to infinity, but the Gaussian function falls back to zero and traces out a bell-shaped curve.

Now let’s consider the content of eqn F.19.

1. Because fis proportional to the range of speeds s, we see that the fraction in the range sincreases in proportion to the width of the range. If at a given speed we double the range of interest (but still ensure that it is narrow), then the fraction of molecules in that range doubles too.

2. Equation F.19 includes a decaying exponential function, the term e^Ms2/2RT. Its presence implies that the fraction of molecules with very high speeds will be very small because e^x2becomes very small when x²is large.

3. The factor M/2RTmultiplyings²in the exponent is large when the molar mass,M,is large, so the exponential factor goes most rapidly toward zero whenMis large. That tells us that heavy molecules are unlikely to be found with very high speeds.

M 2RT

1

1

0 x 0 x

e^−x2

e^−x2

e^−x e^−x

Fig. F.8 The exponential function, e^x, and the bell- shaped Gaussian function, e^x2. Note that both are equal to 1 atx0, but the exponential function rises to infinity as xˆl. The enlargement shows the behavior for x0 in more detail.