# Chapter 5 Gases

## Gas Pressure

Pressure is created by collisions of molecules to the wall

Higher the pressure more collisions b/w molecules/area/time

Pressure is

$P=\frac{F}{A}=\frac{ma}{A}=\frac{mg}{A}=\rho gh$ with F = force, A = area, m = mass, g = gravity, and d = density. In deriving the expression, we used F = ma and m/A = dh.

Unit of pressure = Pascal (abriviated with Pa) = 1 kgm/s2
Barometric pressure = A column with 1 m x 1 m area and a height extending to the stragosphere containing air exerts 101325 Pa.

### Mercury Column Barometer

Torrichelli (1643) used a column of mercury
At sea level height of columen measured = 760 mmHg
Define 760 mmHg = one standard atmosphere or 1 atm

Gravity (g = 9.800665 m/s2) pulls down mercury ⇄ air pressure pushes mercury (d = 13.595 g/cm3) up

Example: A barometer is filled with percholorethylene (dper = 1.62 g/cm3). The liquid height is found to be 6.38 m. What is the barometric pressure in mmHg?

Since PHg = Pper, and expand each P with P = dgh   (make sure to label each with subscripts!)

g dHghHg= g dperhper
dHghHg = dperhper
hHg = dperhper/dHg
hHg = 1.62 g/cm36.38 m/13.595 g/cm3
hHg = 0.760 m = 760 mm = 1.00 atm

### Manometer

Manometer = device to measure gas (well, just a flast with bent tube attached as shown)

If the gas inside the flask exerts the same pressure (Pgas) as atmosheric pressure (Patm), the level of liquid in the tube must be leveled. Let's call it, Δh = 0.

As seen on the graph left

If Pgas < Patm → Δh < 0

This is the case for (a)

If Pgas > Patm → Δh > 0

This is the case for (b)

## Ideal Gas Law (IGL)

4 gas laws combined. 1) Boyle's Law, 2) Charles' Law, Gay-Lussac's Law and 4) Avogadro's Law

• Boyle's Law (1662)
Relationship b/w P and V
Pressure is inversely proportional to Volume $\begin{array}{l}P\propto \frac{1}{V}\\ P=\frac{c}{V}\end{array}$

• Charles' Law (mid 1800's)
Relationship b/w Volume and Temperature
Volume is directly proportional to Temperature. $\begin{array}{l}V\propto T\\ V=cT\end{array}$

When the volume is plotted against temperature in the centigrade scale (a) in the graph, the temperature at which volume becomes 0, is the absolute 0 temperature. So, in (b) we rescale the temperature scale, called absolute temperature scale (the unit is called Kelvin, K).

• Gay-Lussac's Law (aka Amontons' Law): Relationship b/w Pressure and Temperature with constant volume.
Pressure is directly proportional to Temperature. $\begin{array}{l}P\propto T\\ P=cT\end{array}$

• Avogadro's Law (1811) Relationship b/w Volume and quantity of gas molecules (moles).
Volume is directly proportional to a number of moles.

$\begin{array}{l}V\propto n\\ V=cn\end{array}$

At the certain Temperature and Pressure

 2H2 + O2 → 2H2O 2 mol 1 mol 2 mol ↓ ↓ ↓ 2 L 1 L 2 L

• Ideal Gas Law
Combining the four laws
P = c/V      or      PV = c
P = cT
V = cT
V = cn
then
PV = ncT, and let us call c as R to obtain
PV = nRT
with R = 0.08206 Latm/(mol K).

## Ideal Gas Equation

Let's use the IGL eqn to do some calculations!

### Standard Temperature and Pressure (STP)

T = 0 °C = 273.15 K; P = 760 mmHg = 1 atm $PV=nRT$ Solve for V and substitute the numbers in, $V= nRT P = ( 1mol )( 0.08206Latm/molK )( 273.15K ) 1atm =22.4L$

When the volume of one mole of gas is measured, Molar Volume @STP = 22.4 L.

### Initial-Final State Problem

Since R is a constant, the ratio of PV and nT is also constant.

$R=\frac{PV}{nT}$

If we have a gas in a container (initial state, labeled with subscript i), we can modify the condition to the gas (final state, labeled with subscript f). These two states are related by R as

$R=\frac{{P}_{i}{V}_{i}}{{n}_{i}{T}_{i}}=\frac{{P}_{f}{V}_{f}}{{n}_{f}{T}_{f}}$

Example: We have a vessel containing 2 mol of gas at 100°C and at 1.5 atm,as seen in the picture.

If the pressure of the vessel is reduced to 1.00 atm, what is the temperature of the gas?

Since we didn't change the vessel during the experiment, the volume should be the same, and since we didn't open the vessel, number of gas molecules are the same before and after. It means that

$\frac{{P}_{i}}{{T}_{i}}=\frac{{P}_{f}}{{T}_{f}}$

${T}_{f}=\frac{{P}_{f}{T}_{i}}{{P}_{i}}=\frac{\left(1.00atm\right)\left(373K\right)}{1.5atm}=248.67K=-24.5°C$

### Molar Mass

The molar mass, μ with m and n are mass and # of moles, respecitvely, and its rearrangement is, $\begin{array}{l}\mu =\frac{m}{n}\\ n=\frac{m}{\mu }\end{array}$

Substituting above into IGL $\begin{array}{l}PV=\frac{mRT}{\mu }\\ \mu =\frac{mRT}{PV}\end{array}$

Example: A glass vessel weighs 40.1305 g when clean, dry and evacuated. It weighs 138.2410 g when filled with H2O at 25.0°C (ρ = 0.997 g/mL) and weighs 40.2959 g when filled with propylene gas at 740.3 mmHg at 24.0°C. What is the molar mass of propylene?

For propylene gas, we know P and T, so in the above equation for molar mass, we need to obtain m and V.

The mass of the gas is the difference b/w the mass of gass-filled vessel and of the empty vessel, i.e. mgas = mgas+empty - mempty.

The volume of the gas, which equals the volume of the vessel, is obtained by knowing the mass of water in the vessel. The mass of water is converted to volume by its density given.

mH2O = mH2O+empty - mempty
VH2O = mH2O[ 1 mL / 0.997 g ]

Then, substitue these into the expression for molar mass, as
$\mu =\frac{\left(0.1654g\right)\left(0.08206Latm/molK\right)\left(297.15K\right)}{\left(0.9741atm\right)\left(0.09841L\right)}$

μ = 42.08 g/mol

### Gas Density

Rearranging the molar mass equation above, recognizing that m/V is density, ρ, therefore,

$\rho =\frac{m}{V}=\frac{P\mu }{RT}$

Density of air depends on Temperature:

@ 0°C ρ = 1.2922 kg/m3
@15°C ρ = 1.2250 kg/m3
@20°C ρ = 1.2041 kg/m3
@30°C ρ = 1.1164 kg/m3

Example: At what temperature the density of O2 be 1.00 g/L if the pressure is kept at 745 mmHg?

Solve for T in the density equation.

Then, T = (μ P)/(R ρ) = 382.26 K or 109°C

## Gase Stoichiometry

Relating the amount of gas of reactant or product to other species in the reaction by stoichiometry. Notabean: PV = nRT only applies to gas species in the reaction!!

Example: What is the mass of Na(l), in grams, is produced, when 8.00 L of N2(g) at 25.00°C and at 751 mmHg generated in the decomposition of sodium azide (NaN3(s))?

2 NaN3(s) 2Na(l) + 3N2(g)

?mol(N2) = PV/RT = [(0.998158 atm)(8.00L)]/[(0.08206 Latm/molK)(298.15K)] = 0.3263795599 molN2

$?{g}_{Na}=0.32638mo{l}_{{N}_{2}}\left(\frac{2mo{l}_{Na}}{3mo{l}_{{N}_{2}}}\right)\left(\frac{22.9898{g}_{Na}}{1mo{l}_{Na}}\right)=5.002267g=5.00g$

Example: Hydrogen gas can be generated by reaction of Zn metal with hydrochloric acid,

Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g)

a) How many L of H2 would be formed at 742 mmHg an 15°C if 25.5 g of zinc was allowed to react?

b) How many g of Zn would you start with if you wanted to prepare 5.00 L of H2 at 350 mmHg and 30.0°C?

## Partial Pressures

As long as gases are inert (don't react each other), the total # mol, N, is the sum of the # of moles of each molecular species (n1, n2, ...)

Example: A 2.0 L O2(g) and 8.0 L N2(g), each at STP, are mixed. The mixture is compressed to occupy 2.0 L at 298K, what is the total pressure?

?molO2 = PV/RT = (1 atm)(2.0 L)/[(0.9=08206 Latm/molK)(273.15K) = 0.089227 mol(O2

Similarly, nN2 = 0.356909 mol. So, N = n(O2) + n(N2) = 0.446136 mol.

P = nRT/V =(0,446136 mol)(0.08206 Latm/molK)(298K)/2.0 L = 5.454878 atm = 5.4 atm

### Partial Pressure

The sum of the component gas pressures of a gas mixture, is the total pressure of the mixture.

@ constant V and T,
Ptotal = P1 + P2 + P3 + ...
with Pi is the partial pressure of ith gas.

Let's suppose that we have N2 @STP with the volume 22.4 L in one container. It means that we have 1 mol of N2. In antoher container with its volume 22.4 L, we place O2 @STP. So, it also contains 1 mol. Combine these gases, now we have 2 mol of gas, keeping volume V (22.4 L) and temperature T (0°c or 273.15 K) constant. The pressure should then be 2 atm.

So, combined gas has Ptotal = PN2 + PO2

The total pressure is, ${P}_{total}=\frac{{n}_{total}RT}{V}$

Furthermore, we have the relationship ${P}_{total}=\frac{{n}_{total}RT}{V}=\frac{\left({n}_{{N}_{2}}+{n}_{{O}_{2}}\right)RT}{V}=\frac{{n}_{{N}_{2}}RT}{V}+\frac{{n}_{{O}_{2}}RT}{V}={P}_{{N}_{2}}+{P}_{{O}_{2}}$

So, if we take the ratio between PN2 to Ptotal you get,

$\frac{{P}_{{N}_{2}}}{{P}_{total}}=\frac{\frac{{n}_{{N}_{2}}RT}{V}}{\frac{{n}_{total}RT}{V}}=\frac{{n}_{{N}_{2}}}{{n}_{total}}={\chi }_{{N}_{2}}$

The ratio of mol of ith component and the total # of moles is called mole fraction, χi. Then, we can calculate the partial pressure of N2 by knowing χN2 and Ptotal as

${P}_{{N}_{2}}={\chi }_{{N}_{2}}{P}_{total}$

Similar conclusion can be made for VN2 / Vtotal = χN2 if we consider constant P and constant T.

Example: The percent composition by volume of air is 78.08% N2, 20.95% O2, 0.930% Ar, and 0.0360% CO2. What are the partial pressure, if barometric pressure is 748 mmHg (0.984210 atm).

Let's assume at the course of measurement, P, V, and T are constant. Consider Earth as a big container, and its volume doesn't change. Also, if one considers that average pressure at sea level to be fairly constant, we can justify that P, V, and T are constant. Then, we can say that,

$\frac{{V}_{i}}{{V}_{total}}=\frac{{n}_{i}}{{n}_{total}}=\frac{{P}_{i}}{{P}_{total}}$

It means that χs are

speciesχPi
χN20.78080.768 atm
χO20.20950.206 atm
χAr0.00930.00915 atm
χCO20.000360.000354 atm
The total pressure adds up to be 0.984 atm.

Example: A mixture of CS2(g) and excess O2(g) is placed in a 10.0 L reaction vessel at 100.0°C and a pressure of 3.00 atm. The reaction was carried out according to the following equation
CS2(g) + 3O2(g) CO2(g) + 2SO2(g)

After reaction, the temperature was returned to 100.0°C, and the mixture of CO2, SO2 and unreacted O2 is found to have a pressure of 2.40 atm. What is the partial pressure of each gas in the product mixture?

Before the reaction was carried out, we have the values of P, V and T, therefore we can calculate the total number of moles of gases in the reactant mixture from PV = nRT. Similarly, we can calculate the total number of moles of gases in the products and excess O2. Let us call nreact and nprod for the respective number of moles.

Each of the gas molecules is stoichiometrically related. Before the reaction, the total number of moles of all gas species is,

nreact = nCS2 + nO2reacted + nO2excess

We also know the total number of moles of gases after the reaction,

nprod = nCO2 + nSO2 + nO2excess

Since 1 mol of CS2 reacts with 3 mol O2, producing 1 mol of CO2 and 2 mol SO2, we can rewrite the nreact and nprod in terms of number of moles of CS2, such that,

nreact = nCS2 + 3nCS2 + nO2excess = 4nCS2 + nO2excess
nprod = nCS2 + 2nCS2 + nO2excess = 3nCS2 + nO2excess

Now, in the above two equations, we have two unknowns. If we solve for nO2excess in the second equation, we have,

nO2excess = nprod - 3nCS2

Substitute the above into the nreact equation, we get,

nreact = 4nCS2 + nprod - 3nCS2

which is,

nreact = nCS2 + nprod

Therefore, the number of moles of CS2 is now,

nCS2 = nreact - nprod = 0980 mol - 0.784 mol = 0.196 mol

Now that we know the number of moles of CS2, we can relate all other species by stoichiometry in the reaction.

nO2reacted = 3nCS2 = 3 (0.196 mol) = 0.588 mol

nCO2 = nCS2 = 0.196 mol

nSO2 = 2nCS2 = 0.392 mol

nO2excess = nprod - 3nCS2 = 0.784 mol - 3(0.196 mol) = 0.196 mol

From these, we can finally calculate the product mole fractions, χ.

χCO2 = 0.196 mol / 0.784 mol = 0.250

χSO2 = 0.392 mol / 0.784 mol = 0.500

χO2excess = 0.196 mol / 0.784 mol = 0.250

Now, partial pressure, Pi = Ptotal χi , of each gas is
PCO2 = 2.40 atm (0.250) = 0.600 atm

PSO2 = 2.40 atm (0.500) = 1.200 atm

PO2excess = 2.40 atm (0.250) = 0.600 atm

Example: In a 20.0 L flask, 0.776 g He and 3.61 g CO2 are held at 300.0K. Calculate the total pressure and partial pressures of each gas.

We have, V and T. The total number of moles, n, can be calculated by converting the mass to mole. $?mo l He =0.776 g He ( 1mo l He 4.003 g He )=0.19385mo l He ?mo l C O 2 =3.61 g C O 2 ( 1mo l C O 2 42.022 g C O 2 )=0.085907mo l C O 2$ So, the total number of moles is 0.19385 + 0.085907 mol = 0.27975 mol.

$P= nRT V = ( 0.2795mol )( 0.08206Latm/molK )( 300.0K ) 20.0L =0.34435atm$

The partial pressure is obtained from mole fractions. $χ He = n He n total = 0.19385mol 0.27975mol =0.69294 χ C O 2 = n C O 2 n total = 0.085907mol 0.27975mol =0.30708$

From these, we can get partial pressures $P He = χ He P total =( 0.69294 )( 0.34435atm )=0.23854atm=0.238atm P C O 2 = P total − P He =0.34435atm−0.23854atm=0.10574atm=0.106atm$

Example: In bulb A, a mixture of gas, H2O, CO2, and N2 is placed. The valve between bulb A and bulb B is closed. Bulb B is initially empty and is held at a temperature of -70°C. Bulb C is initially empty and is held at -190°C. We also know that CO2 sublimes at -78°C, and N2 boils at -196°C. a) When the valve between A and B is open, then the pressure becomes 219 mmHg. What do bulbs A and B contain? b) How many moles of H2O are in the system? c) Both valves are now open, and the measured pressure is 33.5 mmHg. What do bulbs A, B, and C contain? d) How many moles of N2? e) How many moles of CO2?

a) From the initial condition, we can calculate the total number of moles of gases. $n= PV RT = ( 0.742atm )( 1.00L ) ( 0.08206Latm/molK )( 298.15K ) =0.03033mol$ Since bulb B is held at -70°C all H2O is frozen in bulb B. Therefore, there must be CO2 and N2 are the only ones found in bulbs A and B.

b) To calculate the number of moles of H2O, we need to know the number of moles of gas in bulb A and bulb B separately since these two bulbs have different temperature.

The pressure in both bulbs is $?atm=219mmHg( 1atm 760mmHg )=0.2882atm$ So, in bulb A, $n A = PV RT = ( 0.2882atm )( 1.00L ) ( 0.08206Latm/molK )( 298.15K ) =0.01178mol$ In bulb B, $n B = PV RT = ( 0.2882atm )( 1.00L ) ( 0.08206Latm/molK )( 203.15K ) =0.01729mol$ The total number of gas molecules are then, ntotal = 0.029068 mol. Then, the number of moles of water is 0.03033 mol - 0.029068 mol = 0.001262 mol = 1.26 x 10-3 mol.

c) Now all valves are open. P in each bulb is 33.5 mmHg = 0.04408 atm. Since bulb C is -190°C, CO2 is frozen. So then the contents of each bulb are: bulb A == only N2, bulb B == N2 and H2O, and bulb C == N2 and CO2.

d) As in part b), we can calculate the number of moles of N2 in each bulb, as $n A = PV RT = ( 0.04408atm )( 1.00L ) ( 0.08206Latm/molK )( 298.15K ) =0.0018017mol n B = PV RT = ( 0.04408atm )( 1.00L ) ( 0.08206Latm/molK )( 203.15K ) =0.0026444mol n C = PV RT = ( 0.04408atm )( 1.00L ) ( 0.08206Latm/molK )( 83.15K ) =0.0064602mol$ The total number of moles of gas, which is N2, is 0.010906 mol.

e) The number of moles of CO2 is

$n C O 2 = n original − n H 2 O − n N 2 =0.03033mol−0.001262mol−0.010906mol=0.01816mol$

## Kinetic Theory of Gases

IGL at molecular level -- meaning of kinetic energy (thermal energy)

Our model

• Large # (in the order of 1023) of molecules in constant and random motion
• Molecules collide each other and to the wall to change direction, otherwise straight-line motion
• Gas molecules are sparsely separated, so that the volume is mostly empty space
• No force b/w molecules
• Individual molecules can gain and lose energy, but the total energy is constant

Using this model we can derive the expression for pressure,

$P=\frac{1}{3}\frac{Nm{〈\upsilon 〉}^{2}}{V}$

where N is the # of molecules, m is the mass of molecules <ν> is the average velocity of gas, and V, is the volume of the container.
If you want to see the derivation of the pressure expression, click below!

+ Derivation of P

### Meaning of Kinetic Energy

From Newton's law of motion, we know that kinetic energy, EK = 0.5 mv2, we can derive expression for kinetic energy for gas, ${E}_{K}=\frac{3}{2}RT$ with R = 8.314 J/molK (another way to express IG Constant, also 1 J = 1 kgm2/s). So, for a gas kinetic energy depends only on the temperature. Conversely, temperature measures the how much motion the molecules possess.
Kinetic energy of gas depends only on temperature!
If T → 0, EK → 0

Furthermore, the average speed of the gas molecules can be obtained by rearranging the kinetic energy equation,
$〈v〉={\left(\frac{3RT}{\mu }\right)}^{\frac{1}{2}}$ Therefore, the speed of molecule depends on the molar mass and temperature. Heavier the molecules, slower the speed at given temperature. The speed is higher at higher temperature.

Example: The cruising speed of Boeing 747 jet is 570.0 mi/hr. What temperatures does butane (C4H10) gas, used in cigarette lighter, have the same speed? By the way, don't smoke! It's not good for you!

The volocity in m/s unit is more appropriate than mi/hr. $? m s =570 mi hr ( 1.60934km 1mi )( 10 3 m 1km )( 1hr 60min )( 1min 60s )=158.3km/s$

The following is nearly the same question as above.

Example: The escape velocity required for gas molecules to overcome the earth's gravity and go off to outer space is 1.12 x 103 m/s at 15°C. Calculate the molar mass of a species with that velocity. Would you expect to find He and H2 molecules in the earth's atmosphere? What about argon atom?

### Effusion Rate

• Diffusion - migration of molecules due to random motion
• Effusion - escape of molecules under low pressure

Consider two molecules, lighter A and heavier B, moving in one direction, as shown on left. For a given time interval, distances covered by the molecules differ according to the velocity equation, which depends on molar mass. Lighter molecule travels further for a give time interval, if the two gases are at the same temperature. We can obtain the relationship between the distance traveled is given by the following ratio. $\frac{{\nu }_{A}}{{\nu }_{B}}=\frac{{\left(\frac{3RT}{{\mu }_{A}}\right)}^{\frac{1}{2}}}{{\left(\frac{3RT}{{\mu }_{B}}\right)}^{\frac{1}{2}}}={\left(\frac{{\mu }_{B}}{{\mu }_{A}}\right)}^{\frac{1}{2}}$

Example: When gaseous ammonia and HCl meet, they form a complex NH4Cl, and appear as white cloud. In the tube shown here, concentrated solution of ammonia is place 2 m away from the concentrated HCl. Where do they meet to form white cloud (blue in the figure!)?

Let us define distance covered by HCl as x. Then, the distance covered by NH3 is 2 - x. Since reaching time is the same, the ratio of speeds is the ratio of the distances covered by each. $\frac{{\nu }_{HCl}}{{\nu }_{N{H}_{3}}}=\frac{x}{2-x}={\left(\frac{{\mu }_{N{H}_{3}}}{{\mu }_{HCl}}\right)}^{\frac{1}{2}}$ Then, solve for x, one gets $x=\frac{2{\left(\frac{{\mu }_{N{H}_{3}}}{{\mu }_{HCl}}\right)}^{\frac{1}{2}}}{1+{\left(\frac{{\mu }_{N{H}_{3}}}{{\mu }_{HCl}}\right)}^{\frac{1}{2}}}$ Then, substitute the molar masses in, you get the cloud formation at x = 0.812 m (closer to HCl).

## Deviation from Ideal Behavior

Ideal gas assumptions:
• No volume associated with gas molecules
• No interactions between molecules -- which affects pressure

These assumptions fail at high pressure, as shown below!

The plot is drawn so that compressibility, (PVm/RT), is plotted against pressure, P. Vm is the molar volume; volume of 1 mol of gas.

If the gas behaves as ideal, meaning that it follows IGL, the gas would follow the blue curve. However, real gases behave quite differently at high pressure.

If one takes into account of the excluded volume and molecular interactions, one can obtain the following eqn, called van der Waals' eqn. $( P− a n 2 V 2 )( V−nb )=nRT$ where a and b are the constants specific to gas molecules.