## Substances That Exist as 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/s^{2}

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 mm_{Hg}

Define 760 mm_{Hg} = one standard atmosphere or 1 atm

Gravity (*g* = 9.800665 m/s^{2}) pulls down
mercury ⇄ air pressure pushes mercury (*d* = 13.595
g/cm^{3}) up

** Example:** A barometer is filled with percholorethylene (

*d*

_{per}= 1.62 g/cm

^{3}). The liquid height is found to be 6.38 m. What is the barometric pressure in mm

_{Hg}?

Since *P _{Hg} = P_{per}, and expand each P with P = dgh (make sure to label each with subscripts!)*

*g d*

d

h

h

h

_{Hg}h_{Hg}= g d_{per}h_{per}d

_{Hg}h_{Hg}= d_{per}h_{per}h

_{Hg}= d_{per}h_{per}/d_{Hg}h

_{Hg}= 1.62 g/cm^{3}6.38 m/13.595 g/cm^{3}h

_{Hg}= 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 (P_{gas}) as atmosheric pressure (P_{atm}), the level of liquid in the tube must be leveled. Let's call it, Δ*h* = 0.

As seen on the graph left

If P_{gas} < P_{atm} → Δh < 0

This is the case for (a)

If P_{gas} > P_{atm} → 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

Relationship b/w P and V

Pressure is inversely proportional to Volume $$\begin{array}{l}P\propto \frac{1}{V}\\ P=\frac{c}{V}\end{array}$$

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*).

Pressure is directly proportional to Temperature. $$\begin{array}{l}P\propto T\\ P=cT\end{array}$$

Volume is directly proportional to a number of moles.

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

2H_{2} | + | O_{2} | → | 2H_{2}O |

2 mol | 1 mol | 2 mol | ||

↓ | ↓ | ↓ | ||

2 L | 1 L | 2 L |

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*

*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 mm

_{Hg}= 1 atm $$PV=nRT$$ Solve for

*V*and substitute the numbers in, $$V=\frac{nRT}{P}=\frac{\left(1mol\right)\left(0.08206Latm/molK\right)\left(273.15K\right)}{1atm}=22.4L$$

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

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

$${T}_{f}=\frac{{P}_{f}{T}_{i}}{{P}_{i}}=\frac{(1.00atm)(373K)}{1.5atm}=248.67K=-24.5\xb0C$$

### 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 H

_{2}O at 25.0°C (ρ = 0.997 g/mL) and weighs 40.2959 g when filled with propylene gas at 740.3 mm

_{Hg}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. m _{gas} = m_{gas+empty} - m_{empty}*.

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.

*m _{H2O} = m_{H2O+empty} - m_{empty}*

*V*

_{H2O}= m_{H2O}[ 1 mL / 0.997 g ]
Then, substitue these into the expression for molar mass, as

$$\mu =\frac{(0.1654g)(0.08206Latm/molK)(297.15K)}{(0.9741atm)(0.09841L)}$$

μ = 42.08 g/mol

### Gas Density

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

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

@ 0°C ρ = 1.2922 kg/m^{3}

@15°C ρ = 1.2250 kg/m^{3}

@20°C ρ = 1.2041 kg/m^{3}

@30°C ρ = 1.1164 kg/m^{3}

**At what temperature the density of O**

*Example:*_{2}be 1.00 g/L if the pressure is kept at 745 mm

_{Hg}?

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!!

**What is the mass of Na(**

*Example:**l*), in grams, is produced, when 8.00 L of N

_{2}(

*g*) at 25.00°C and at 751 mm

_{Hg}generated in the decomposition of sodium azide (NaN

_{3}(

*s*))?

_{3}(

*s*) → 2Na(

*l*) + 3N

_{2}(

*g*)

_{2}) = PV/RT = [(0.998158 atm)(8.00L)]/[(0.08206 Latm/molK)(298.15K)] = 0.3263795599 mol

_{N2}

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

*Example:**s*) + 2HCl(

*aq*) → ZnCl

_{2}(

*aq*) + H

_{2}(

*g*)

a) How many L of H_{2} 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 H_{2} 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 (n_{1}, n

_{2}, ...)

**A 2.0 L O**

*Example:*_{2}(g) and 8.0 L N

_{2}(g), each at STP, are mixed. The mixture is compressed to occupy 2.0 L at 298K, what is the total pressure?

_{O2}= PV/RT = (1 atm)(2.0 L)/[(0.9=08206 Latm/molK)(273.15K) = 0.089227 mol(O

_{2}

Similarly, n_{N2} = 0.356909 mol. So, N = n(O_{2}) + n(N_{2}) = 0.446136 mol.

### Partial Pressure

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

*P*

_{total}= P_{1}+ P_{2}+ P_{3}+ ...with

*P*is the partial pressure of i

_{i}^{th}gas.

Let's suppose that we have N_{2} @STP with the volume 22.4 L in one container. It means that we have 1 mol of N_{2}. In antoher container with its volume 22.4 L, we place O_{2} @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 *P _{total} = P_{N2} + P_{O2}*

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 *P _{N2}* to

*P*you get,

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

The ratio of mol of i^{th} component and the total # of moles is called mole fraction,* χ _{i}*. Then, we can calculate the partial pressure of N

_{2}by knowing

*χ*and

_{N2}*P*as

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

Similar conclusion can be made for *V _{N2} / V_{total} = χ_{N2}* if we consider constant

*P*and constant

*T*.

*The percent composition by volume of air is 78.08% N*

**Example:**_{2}, 20.95% O

_{2}, 0.930% Ar, and 0.0360% CO

_{2}. What are the partial pressure, if barometric pressure is 748 mm

_{Hg}(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

species | χ | P_{i} |
---|---|---|

χ_{N2} | 0.7808 | 0.768 atm |

χ_{O2} | 0.2095 | 0.206 atm |

χ_{Ar} | 0.0093 | 0.00915 atm |

χ_{CO2} | 0.00036 | 0.000354 atm |

**A mixture of CS**

*Example:*_{2}(

*g*) and excess O

_{2}(

*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

_{2}(

*g*) + 3O

_{2}(

*g*) → CO

_{2}(

*g*) + 2SO

_{2}(

*g*)

After reaction, the temperature was returned to 100.0°C, and the mixture
of CO_{2}, SO_{2} and unreacted O_{2} 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 O_{2}.
Let us call *n _{react}* and

*n*for the respective number of moles.

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

*n*

_{react}= n_{CS2}+ n_{O2}^{reacted}+ n_{O2}^{excess}We also know the total number of moles of gases after the reaction,

*n*

_{prod}= n_{CO2}+ n_{SO2}+ n_{O2}^{excess}
Since 1 mol of CS_{2} reacts with 3 mol O_{2}, producing
1 mol of CO_{2} and 2 mol SO_{2}, we can rewrite the
*n _{react}* and

*n*in terms of number of moles of CS

_{prod}_{2}, such that,

*n*

_{react}= n_{CS2}+ 3n_{CS2}+ n_{O2}^{excess}= 4n_{CS2}+ n_{O2}^{excess}*n*

_{prod}= n_{CS2}+ 2n_{CS2}+ n_{O2}^{excess}= 3n_{CS2}+ n_{O2}^{excess}
Now, in the above two equations, we have two unknowns. If we solve for
*n _{O2}^{excess}* in the second equation,
we have,

*n*

_{O2}^{excess}= n_{prod}- 3n_{CS2}
Substitute the above into the *n _{react}* equation, we get,

*n*

_{react}= 4n_{CS2}+ n_{prod}- 3n_{CS2}which is,

*n*

_{react}= n_{CS2}+ n_{prod}
Therefore, the number of moles of CS_{2} is now,

*n*= 0980 mol - 0.784 mol = 0.196 mol

_{CS2}= n_{react}- n_{prod}
Now that we know the number of moles of CS_{2},
we can relate all other species by stoichiometry in the reaction.

*n*= 3 (0.196 mol) = 0.588 mol

_{O2}^{reacted}= 3n_{CS2 }
*n _{CO2} = n_{CS2}* = 0.196 mol

*n _{SO2} = 2n_{CS2}* = 0.392 mol

*n _{O2}^{excess} = n_{prod} -
3n_{CS2}* = 0.784 mol - 3(0.196 mol) = 0.196 mol

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

*χ*= 0.196 mol / 0.784 mol = 0.250

_{CO2}
*χ _{SO2}* = 0.392 mol / 0.784 mol = 0.500

*χ _{O2}^{excess}* =
0.196 mol / 0.784 mol = 0.250

*P*, of each gas is

_{i}= P_{total}χ_{i}*P*= 2.40 atm (0.250) = 0.600 atm

_{CO2}
*P _{SO2}* = 2.40 atm (0.500) = 1.200 atm

*P _{O2}^{excess}* = 2.40 atm (0.250) = 0.600 atm

**In a 20.0 L flask, 0.776 g He and 3.61 g CO**

*Example:*_{2}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.
$$\begin{array}{l}?mo{l}_{He}=0.776{g}_{He}\left(\frac{1mo{l}_{He}}{4.003{g}_{He}}\right)=0.19385mo{l}_{He}\\ ?mo{l}_{C{O}_{2}}=3.61{g}_{C{O}_{2}}\left(\frac{1mo{l}_{C{O}_{2}}}{42.022{g}_{C{O}_{2}}}\right)=0.085907mo{l}_{C{O}_{2}}\end{array}$$
So, the total number of moles is 0.19385 + 0.085907 mol = 0.27975 mol.

$$P=\frac{nRT}{V}=\frac{\left(0.2795mol\right)\left(0.08206Latm/molK\right)\left(300.0K\right)}{20.0L}=0.34435atm$$

The partial pressure is obtained from mole fractions. $$\begin{array}{l}{\chi}_{He}=\frac{{n}_{He}}{{n}_{total}}=\frac{0.19385mol}{0.27975mol}=0.69294\\ {\chi}_{C{O}_{2}}=\frac{{n}_{C{O}_{2}}}{{n}_{total}}=\frac{0.085907mol}{0.27975mol}=0.30708\end{array}$$

From these, we can get partial pressures $$\begin{array}{l}{P}_{He}={\chi}_{He}{P}_{total}=\left(0.69294\right)\left(0.34435atm\right)=0.23854atm=0.238atm\\ {P}_{C{O}_{2}}={P}_{total}-{P}_{He}=0.34435atm-0.23854atm=0.10574atm=0.106atm\end{array}$$

**In bulb A, a mixture of gas, H**

*Example:*_{2}O, CO

_{2}, and N

_{2}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 CO

_{2}sublimes at -78°C, and N

_{2}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 H

_{2}O 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 N

_{2}? e) How many moles of CO

_{2}?

a) From the initial condition, we can calculate the total number of moles
of gases.
$$n=\frac{PV}{RT}=\frac{\left(0.742atm\right)\left(1.00L\right)}{\left(0.08206Latm/molK\right)\left(298.15K\right)}=0.03033mol$$
Since bulb B is held at -70°C all H_{2}O is frozen in bulb B.
Therefore, there must be CO_{2} and N_{2} are the only
ones found in bulbs A and B.

b) To calculate the number of moles of H_{2}O, 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
$$?atm=219mmHg\left(\frac{1atm}{760mmHg}\right)=0.2882atm$$
So, in bulb A,
$${n}_{A}=\frac{PV}{RT}=\frac{\left(0.2882atm\right)\left(1.00L\right)}{\left(0.08206Latm/molK\right)\left(298.15K\right)}=0.01178mol$$
In bulb B,
$${n}_{B}=\frac{PV}{RT}=\frac{\left(0.2882atm\right)\left(1.00L\right)}{\left(0.08206Latm/molK\right)\left(203.15K\right)}=0.01729mol$$
The total number of gas molecules are then, *n*_{total} =
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, CO_{2} is frozen. So then the
contents of each bulb are: bulb A == only N_{2}, bulb B == N_{2}
and H_{2}O, and bulb C == N_{2} and CO_{2}.

d) As in part b), we can calculate the number of moles of N_{2} in each
bulb, as
$$\begin{array}{l}{n}_{A}=\frac{PV}{RT}=\frac{\left(0.04408atm\right)\left(1.00L\right)}{\left(0.08206Latm/molK\right)\left(298.15K\right)}=0.0018017mol\\ {n}_{B}=\frac{PV}{RT}=\frac{\left(0.04408atm\right)\left(1.00L\right)}{\left(0.08206Latm/molK\right)\left(203.15K\right)}=0.0026444mol\\ {n}_{C}=\frac{PV}{RT}=\frac{\left(0.04408atm\right)\left(1.00L\right)}{\left(0.08206Latm/molK\right)\left(83.15K\right)}=0.0064602mol\end{array}$$
The total number of moles of gas, which is N_{2}, is 0.010906 mol.

e) The number of moles of CO_{2} 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

^{23}) of molecules in constant and random motion

Using this model we can derive the expression for pressure,

$$P=\frac{1}{3}\frac{Nm{\langle \upsilon \rangle}^{2}}{V}$$

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!

*P*

### Meaning of Kinetic Energy

From Newton's law of motion, we know that kinetic energy,*E*= 0.5 mv

_{K}^{2}, 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 kgm

^{2}/s

If T → 0,

*E*→ 0

_{K}
Furthermore, the average speed of the gas molecules can be obtained by rearranging the kinetic energy equation,

$$\langle v\rangle ={\left(\frac{3RT}{\mu}\right)}^{\frac{1}{2}}$$
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.

*The cruising speed of Boeing 747 jet is 570.0 mi/hr. What temperatures does butane (C*

**Example:**_{4}H

_{10}) 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. $$?\frac{m}{s}=570\frac{mi}{hr}\left(\frac{1.60934km}{1mi}\right)\left(\frac{{10}^{3}m}{1km}\right)\left(\frac{1hr}{60\mathrm{min}}\right)\left(\frac{1\mathrm{min}}{60s}\right)=158.3km/s$$

The following is nearly the same question as above.

*The escape velocity required for gas molecules to overcome the earth's gravity and go off to outer space is 1.12 x 10*

**Example:**^{3}m/s at 15°C. Calculate the molar mass of a species with that velocity. Would you expect to find He and H

_{2}molecules in the earth's atmosphere? What about argon atom?

### Effusion Rate

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.
$$\frac{{\nu}_{A}}{{\nu}_{B}}=\frac{{\left(\frac{3RT}{{\mu}_{A}}\right)}^{\frac{1}{2}}}{{\left(\frac{3RT}{{\mu}_{B}}\right)}^{\frac{1}{2}}}={\left(\frac{{\mu}_{B}}{{\mu}_{A}}\right)}^{\frac{1}{2}}$$

*When gaseous ammonia and HCl meet, they form a complex NH*

**Example:**_{4}Cl, 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!)?

*x*. Then, the distance covered by NH

_{3}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:These assumptions fail at high pressure, as shown below!

The plot is drawn so that *compressibility*, (*PV _{m}/RT*),
is plotted against pressure,

*P*.

*V*is the molar volume; volume of 1 mol of gas.

_{m}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.
$$\left(P-\frac{a{n}^{2}}{{V}^{2}}\right)\left(V-nb\right)=nRT$$
where *a* and *b* are the constants specific to gas molecules.