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Chapter 5 States Of Matter
We previously learned about properties of individual atoms and molecules. However, many observable characteristics of matter are bulk properties, arising from collections of a large number of particles. For example, a single water molecule doesn't boil, but a collection of them (liquid water) does. Matter exists in different states – solid, liquid, and gas – and these states have distinct physical properties, even if the chemical composition remains the same (like water existing as ice, liquid water, or steam). The state of matter depends on the balance between the energy of the particles (thermal energy) and how they interact with each other (intermolecular forces).
While chemical properties are generally independent of physical state, the rate of chemical reactions can be affected by it. Understanding the physical laws governing the behavior of matter in different states, especially gases and liquids, is essential in chemistry.
To understand the states of matter, we need to consider the nature of intermolecular forces (attractions and repulsions between particles) and thermal energy (energy due to particle motion).
Intermolecular Forces
Intermolecular forces are the attractive and repulsive forces acting between atoms and molecules. This term specifically excludes intramolecular forces (like covalent or ionic bonds) that hold atoms together within a molecule or ion, and electrostatic forces between discrete ions.
Attractive intermolecular forces are collectively known as van der Waals forces, named after Johannes van der Waals, who used them to explain the non-ideal behavior of real gases. Van der Waals forces vary in strength and include:
- Dispersion forces (London forces)
- Dipole-dipole forces
- Dipole-induced dipole forces
Hydrogen bonding is a particularly strong type of dipole-dipole interaction, often considered a separate category because it only involves hydrogen bonded to highly electronegative atoms (like N, O, F, and sometimes Cl).
Forces between an ion and a polar molecule (dipole) are called ion-dipole forces; these are not considered van der Waals forces.
Dispersion Forces Or London Forces
These forces are present between all atoms and molecules, including nonpolar ones. Even in electrically symmetrical atoms or nonpolar molecules, the electron cloud can fluctuate momentarily, leading to a temporary or instantaneous dipole. This instantaneous dipole can then induce a temporary dipole in a neighboring atom or molecule by distorting its electron cloud.
These momentary dipoles attract each other. This weak, short-range attraction is known as a London force or dispersion force. These forces are always attractive and their interaction energy is inversely proportional to the sixth power of the distance ($1/r^6$) between the particles. They are significant only at very short distances ($\approx 500$ pm) and their strength depends on the polarisability of the particle (how easily its electron cloud can be distorted). Larger molecules or atoms are generally more polarisable and experience stronger dispersion forces.
Dipole - Dipole Forces
These forces exist between molecules that have permanent dipoles (polar molecules). Polar molecules have a permanent separation of partial positive and partial negative charges due to differences in electronegativity between bonded atoms and the molecule's geometry.
The partially positively charged end of one polar molecule is attracted to the partially negatively charged end of a neighboring polar molecule. This interaction is stronger than London forces because permanent charges are involved, but weaker than ionic interactions because the charges are only partial. Dipole-dipole interaction energy depends on the distance ($r$) and orientation of the dipoles. For stationary polar molecules, energy is proportional to $1/r^3$; for rotating molecules, it's proportional to $1/r^6$. Polar molecules also experience London forces; the total intermolecular force is a cumulative effect.
Dipole–Induced Dipole Forces
These attractive forces occur between a polar molecule with a permanent dipole and a nonpolar molecule. The permanent dipole of the polar molecule can induce a temporary dipole in the nonpolar molecule by distorting its electron cloud.
The interaction energy is proportional to $1/r^6$. The strength of this force depends on the magnitude of the permanent dipole and the polarisability of the nonpolar molecule. Larger nonpolar molecules are more polarisable, leading to stronger induced dipoles and interactions. Like dipole-dipole interactions, these forces contribute to the total intermolecular forces in mixtures of polar and nonpolar substances.
Hydrogen Bond
Hydrogen bonding is a particularly strong type of dipole-dipole interaction that occurs when a hydrogen atom is covalently bonded to a highly electronegative atom, typically N, O, or F. The H atom becomes partially positive ($\delta+$) and is attracted to a lone pair of electrons on an electronegative atom ($\delta-$) in a neighbouring molecule (or sometimes within the same molecule). Although typically associated with N, O, and F, Chlorine may also participate in weaker hydrogen bonds.
Hydrogen bond energies are significant (10-100 kJ mol$^{-1}$) compared to typical van der Waals forces, playing a crucial role in the properties of many substances like water, proteins, and DNA. We studied this in more detail in Unit 4.
In addition to attractive forces, molecules also experience strong repulsive forces when they are brought very close together. These repulsions arise from the overlap of electron clouds and between the nuclei. The magnitude of these repulsive forces increases rapidly as the distance between molecules decreases. Repulsive forces explain why liquids and solids are much less compressible than gases.
Thermal Energy
Thermal energy is the energy a substance possesses due to the constant motion of its atoms or molecules. This energy is directly proportional to the absolute temperature of the substance. It is a measure of the average kinetic energy of the particles and is responsible for their movement, referred to as thermal motion.
Higher temperatures mean higher thermal energy and faster particle motion.
Intermolecular Forces Vs Thermal Interactions
The state of matter of a substance is determined by the delicate balance between the attractive intermolecular forces (which tend to hold particles together) and the thermal energy (which tends to keep them apart through motion).
- In solids, intermolecular forces are strong and thermal energy is low. Particles are held in fixed positions, vibrating about those positions.
- In liquids, intermolecular forces are still significant, but thermal energy is higher than in solids. Particles are close together but can move past each other.
- In gases, thermal energy is high, and intermolecular forces are relatively weak compared to the kinetic energy of the particles. Particles are far apart and move randomly.
If intermolecular forces are very weak, the substance exists as a gas unless temperature is significantly lowered to reduce thermal energy and allow weak attractions to hold particles together (liquefaction/solidification). Compression alone is not sufficient to liquefy gases unless thermal energy is also reduced, because even when molecules are close, high kinetic energy can overcome attractive forces.
Relative predominance of forces/energy in different states:
- Gases: Thermal energy >> Intermolecular forces
- Liquids: Thermal energy $\approx$ Intermolecular forces
- Solids: Intermolecular forces >> Thermal energy
The Gaseous State
The gaseous state is the simplest state of matter, characterized by weak intermolecular forces and high particle kinetic energy. Air, the atmosphere we live in, is a mixture of gases.
Only a limited number of elements exist as gases under normal conditions (room temperature and pressure): Hydrogen (H$_2$), Nitrogen (N$_2$), Oxygen (O$_2$), Fluorine (F$_2$), Chlorine (Cl$_2$), and the noble gases (He, Ne, Ar, Kr, Xe, Rn).
Physical properties of gases:
- Highly compressible.
- Exert pressure equally in all directions.
- Have much lower density compared to solids and liquids.
- Have no fixed volume or shape; they fill and assume the shape of their container.
- Mix completely and evenly with other gases in all proportions (diffusion).
The simplicity of gases allows their behavior to be described by general laws relating measurable properties: pressure ($p$), volume ($V$), temperature ($T$), and amount ($n$). These relationships are known as the gas laws.
The Gas Laws
The fundamental gas laws describe the relationships between the pressure, volume, temperature, and amount of a gas, based on centuries of experimental observations. Key contributors include Robert Boyle, Jacques Charles, Joseph Louis Gay-Lussac, and Amedeo Avogadro.
Boyle’s Law (Pressure - Volume Relationship)
Formulated by Robert Boyle in 1662, this law states that at constant temperature ($T$) and for a fixed amount ($n$) of gas, the pressure ($p$) of the gas is inversely proportional to its volume ($V$).
$p \propto \frac{1}{V}$ (at constant $T, n$)
This can be written as $pV = k_1$, where $k_1$ is a constant. This means that for a fixed amount of gas at a constant temperature, the product of pressure and volume is always constant.
If the conditions of a gas change from ($p_1, V_1$) to ($p_2, V_2$) at constant $T$ and $n$, then:
$p_1V_1 = p_2V_2 = k_1$
Graphical representations of Boyle's law:
Plot of $p$ vs $V$ gives a hyperbola (isotherm). Different temperatures result in different curves, with higher curves representing higher temperatures (as $k_1$ increases with $T$).
Plot of $p$ vs $1/V$ gives a straight line passing through the origin. At very high pressures, real gases deviate, and the line is no longer perfectly straight.
Boyle's law demonstrates the high compressibility of gases. The density ($d$) of a fixed mass ($m$) of gas is related to volume by $d = m/V$. Since $V = k_1/p$, substituting gives $d = m / (k_1/p) = (m/k_1) \times p = k'p$. Thus, at constant temperature, the density of a fixed amount of gas is directly proportional to its pressure.
Problem 5.1. A balloon is filled with hydrogen at room temperature. It will burst if pressure exceeds 0.2 bar. If at 1 bar pressure the gas occupies 2.27 L volume, upto what volume can the balloon be expanded ?
Answer:
We can use Boyle's Law since the temperature and the amount of hydrogen gas in the balloon are constant.
Boyle's Law states: $p_1V_1 = p_2V_2$.
Given initial conditions:
- $p_1 = 1$ bar
- $V_1 = 2.27$ L
Given the maximum pressure the balloon can withstand before bursting:
- $p_{\text{max}} = 0.2$ bar. If the pressure exceeds 0.2 bar, it bursts. So, the maximum allowable pressure is 0.2 bar.
- We want to find the maximum volume ($V_{\text{max}}$) the balloon can reach without bursting, which corresponds to the pressure of 0.2 bar. Let $p_2 = 0.2$ bar and $V_2 = V_{\text{max}}$.
Using Boyle's Law equation: $p_1V_1 = p_2V_2$.
$(1 \text{ bar}) \times (2.27 \text{ L}) = (0.2 \text{ bar}) \times V_2$.
To find $V_2$, rearrange the equation:
$V_2 = \frac{(1 \text{ bar}) \times (2.27 \text{ L})}{0.2 \text{ bar}}$.
$V_2 = \frac{2.27}{0.2} \text{ L}$.
$V_2 = 11.35 \text{ L}$.
This is the volume the balloon reaches at the maximum allowable pressure (0.2 bar). If the volume exceeds this value, the pressure would drop below 0.2 bar (assuming constant T and n), but the problem states it bursts if pressure *exceeds* 0.2 bar, implying 0.2 bar is the limit. So, the balloon can be expanded up to, but not exceeding, this volume.
The balloon can be expanded up to a volume of 11.35 L.
Charles’ Law (Temperature - Volume Relationship)
Studies on hot air balloons led Charles and Gay-Lussac to investigate the relationship between the volume and temperature of gases. Charles's Law states that for a fixed amount ($n$) of gas at constant pressure ($p$), the volume ($V$) is directly proportional to its absolute temperature ($T$).
Early observations showed that for every degree Celsius rise in temperature, the volume increased by $1/273.15$ of its volume at 0$^\circ$C. This led to the definition of the Kelvin temperature scale (Absolute scale), where $T(\text{K}) = t(^\circ\text{C}) + 273.15$. Absolute zero (0 K or -273.15$^\circ$C) is the hypothetical temperature at which an ideal gas would have zero volume.
Mathematically, Charles's Law is expressed as:
$V \propto T$ (at constant $p, n$)
or $V/T = k_2$, where $k_2$ is a constant.
For a gas changing from ($V_1, T_1$) to ($V_2, T_2$) at constant $p$ and $n$:
$\frac{V_1}{T_1} = \frac{V_2}{T_2} = k_2$
A graph of $V$ vs $T$ (in Kelvin) is a straight line passing through the origin. A graph of $V$ vs $t$ (in Celsius) is a straight line that intercepts the temperature axis at -273.15$^\circ$C (absolute zero). Each line on a $V$ vs $T$ graph at constant pressure is called an isobar.
All gases obey Charles's Law approximately at low pressures and high temperatures.
Problem 5.2. On a ship sailing in Pacific Ocean where temperature is 23.4°C, a balloon is filled with 2 L air. What will be the volume of the balloon when the ship reaches Indian ocean, where temperature is 26.1°C ?
Answer:
We can use Charles's Law here, assuming the amount of air in the balloon and the pressure remain constant (though pressure might change slightly, we assume it's constant for this problem based on using Charles's Law).
Charles's Law states: $\frac{V_1}{T_1} = \frac{V_2}{T_2}$. Temperatures must be in Kelvin.
Given initial conditions in Pacific Ocean:
- Volume $V_1 = 2$ L
- Temperature $t_1 = 23.4^\circ$C. Convert to Kelvin: $T_1 = 23.4 + 273.15 = 296.55$ K. (Text uses 273, let's stick to 273.15 for accuracy, but mention the text's usage). Text uses $23.4+273 = 296.4$ K. Let's use 296.4 K.
Given final conditions in Indian Ocean:
- Temperature $t_2 = 26.1^\circ$C. Convert to Kelvin: $T_2 = 26.1 + 273.15 = 299.25$ K. Text uses $26.1+273 = 299.1$ K. Let's use 299.1 K.
- We need to find the final volume $V_2$.
Using Charles's Law equation: $\frac{V_1}{T_1} = \frac{V_2}{T_2}$.
Rearranging to solve for $V_2$:
$V_2 = V_1 \times \frac{T_2}{T_1}$.
Substitute the given values:
$V_2 = 2 \text{ L} \times \frac{299.1 \text{ K}}{296.4 \text{ K}}$.
$V_2 = 2 \text{ L} \times 1.009109$
$V_2 \approx 2.0182$ L.
Rounding to three significant figures (based on 2 L):
$V_2 \approx 2.02$ L.
The volume of the balloon when the ship reaches the Indian Ocean will be approximately 2.02 L.
Gay Lussac’s Law (Pressure- Temperature Relationship)
Joseph Gay-Lussac's Law states that for a fixed amount ($n$) of gas at constant volume ($V$), the pressure ($p$) is directly proportional to its absolute temperature ($T$).
$p \propto T$ (at constant $V, n$)
or $p/T = k_3$, where $k_3$ is a constant.
For a gas changing from ($p_1, T_1$) to ($p_2, T_2$) at constant $V$ and $n$:
$\frac{p_1}{T_1} = \frac{p_2}{T_2} = k_3$
A graph of $p$ vs $T$ (in Kelvin) at constant volume is a straight line passing through the origin. Each line is called an isochore.
Avogadro Law (Volume - Amount Relationship)
Amedeo Avogadro's Law (1811) states that equal volumes of all gases, under the same conditions of temperature and pressure, contain the same number of molecules. This means that at constant temperature and pressure, the volume ($V$) of a gas is directly proportional to the amount of gas (number of moles, $n$).
$V \propto n$ (at constant $T, p$)
or $V = k_4 n$, where $k_4$ is a constant.
One mole of any gas contains $6.022 \times 10^{23}$ molecules (Avogadro's number). According to Avogadro's law, one mole of any gas should occupy the same volume at standard temperature and pressure (STP).
Historically, STP was defined as 0$^\circ$C (273.15 K) and 1 atm (101.325 kPa). The molar volume of an ideal gas at this STP is approximately 22.414 L/mol.
The current IUPAC standard STP is 273.15 K (0$^\circ$C) and 1 bar (100 kPa). The molar volume of an ideal gas at this STP is approximately 22.711 L/mol.
Gas | Molar Volume (L/mol) |
Argon | 22.37 |
Carbon dioxide | 22.54 |
Dinitrogen | 22.69 |
Dioxygen | 22.69 |
Dihydrogen | 22.72 |
Ideal gas | 22.71 |
The number of moles ($n$) can be calculated from the mass ($m$) and molar mass ($M$) of the gas: $n = m/M$. Substituting this into Avogadro's law ($V = k_4 n$) gives $V = k_4 (m/M)$, or $M = k_4 (m/V) = k_4 d$. This shows that at constant $T$ and $p$, the molar mass of a gas is directly proportional to its density.
An ideal gas is a hypothetical gas that strictly obeys Boyle's, Charles's, and Avogadro's laws under all conditions. In an ideal gas model, intermolecular forces are assumed to be absent, and the volume of the gas particles themselves is considered negligible. Real gases approximate ideal gas behavior only under specific conditions (usually low pressure and high temperature) where intermolecular forces are minimal. Under other conditions, real gases deviate from ideal behavior.
Ideal Gas Equation
The three gas laws (Boyle's, Charles's, and Avogadro's) can be combined into a single equation known as the Ideal Gas Equation. It relates pressure ($p$), volume ($V$), temperature ($T$), and amount ($n$) of an ideal gas.
From Boyle's Law: $V \propto 1/p$ (constant $T, n$)
From Charles's Law: $V \propto T$ (constant $p, n$)
From Avogadro's Law: $V \propto n$ (constant $T, p$)
Combining these proportionalities: $V \propto \frac{nT}{p}$
Introducing a proportionality constant, $R$, we get: $V = R \frac{nT}{p}$
Rearranging gives the ideal gas equation:
$pV = nRT$
$R$ is called the gas constant. It is a universal constant, meaning it has the same value for all ideal gases. Its value depends on the units used for $p$, $V$, and $T$.
$R = \frac{pV}{nT}$
Value of $R$ at current STP (273.15 K and 1 bar) where molar volume is 22.710981 L/mol:
$R = \frac{(1 \text{ bar}) \times (22.710981 \text{ L})}{ (1 \text{ mol}) \times (273.15 \text{ K})} \approx 0.08314 \text{ bar L K}^{-1} \text{ mol}^{-1}$.
Value of $R$ at older STP (273.15 K and 1 atm = 1.01325 bar) where molar volume is 22.414 L/mol:
$R = \frac{(1 \text{ atm}) \times (22.414 \text{ L})}{ (1 \text{ mol}) \times (273.15 \text{ K})} \approx 0.082057 \text{ L atm K}^{-1} \text{ mol}^{-1}$.
In SI units ($p$ in Pa, $V$ in m$^3$, $T$ in K, $n$ in mol): 1 bar = $10^5$ Pa, 1 L = 1 dm$^3 = 10^{-3}$ m$^3$.
$R = \frac{(10^5 \text{ Pa}) \times (22.710981 \times 10^{-3} \text{ m}^3)}{ (1 \text{ mol}) \times (273.15 \text{ K})} \approx 8.314 \text{ Pa m}^3 \text{ K}^{-1} \text{ mol}^{-1}$.
Since Pa m$^3$ is equivalent to Joules (J), $R = 8.314$ J K$^{-1}$ mol$^{-1}$.
The ideal gas equation is also called the equation of state for an ideal gas because it describes the relationship between the variables defining the state of the gas. If the state of a fixed amount of gas changes from ($p_1, V_1, T_1$) to ($p_2, V_2, T_2$), the ideal gas equation can be applied to both states:
$p_1V_1 = nRT_1 \implies \frac{p_1V_1}{T_1} = nR$
$p_2V_2 = nRT_2 \implies \frac{p_2V_2}{T_2} = nR$
Since $n$ and $R$ are constant, we get the Combined Gas Law:
$\frac{p_1V_1}{T_1} = \frac{p_2V_2}{T_2}$
This equation is useful for solving problems where the state of a gas changes.
Problem 5.3. At 25°C and 760 mm of Hg pressure a gas occupies 600 mL volume. What will be its pressure at a height where temperature is 10°C and volume of the gas is 640 mL.
Answer:
We are given initial and final conditions for a fixed amount of gas and need to find the final pressure. We can use the Combined Gas Law equation: $\frac{p_1V_1}{T_1} = \frac{p_2V_2}{T_2}$. Temperatures must be in Kelvin.
Given initial conditions:
- $p_1 = 760$ mm Hg
- $V_1 = 600$ mL
- $t_1 = 25^\circ$C. Convert to Kelvin: $T_1 = 25 + 273.15 = 298.15$ K. Text uses 273, so $T_1 = 25+273 = 298$ K. Let's use 298 K.
Given final conditions:
- $V_2 = 640$ mL
- $t_2 = 10^\circ$C. Convert to Kelvin: $T_2 = 10 + 273.15 = 283.15$ K. Text uses 273, so $T_2 = 10+273 = 283$ K. Let's use 283 K.
- We need to find the final pressure $p_2$.
Using the Combined Gas Law equation: $\frac{p_1V_1}{T_1} = \frac{p_2V_2}{T_2}$.
Rearranging to solve for $p_2$:
$p_2 = \frac{p_1V_1T_2}{T_1V_2}$.
Substitute the given values:
$p_2 = \frac{(760 \text{ mm Hg}) \times (600 \text{ mL}) \times (283 \text{ K})}{(298 \text{ K}) \times (640 \text{ mL})}$.
$p_2 = \frac{760 \times 600 \times 283}{298 \times 640}$ mm Hg.
$p_2 = \frac{129168000}{190720}$ mm Hg.
$p_2 \approx 677.27$ mm Hg.
Rounding to three significant figures (based on 600 mL and 640 mL):
$p_2 \approx 677$ mm Hg.
The pressure of the gas at the new height will be approximately 677 mm Hg.
Density And Molar Mass Of A Gaseous Substance
The ideal gas equation can be used to relate density ($d$) and molar mass ($M$) of a gas.
From $pV = nRT$, we can write $n/V = p/RT$.
Since $n = m/M$, we have $(m/M)/V = p/RT$.
$(m/V)/M = p/RT$.
Since $m/V = d$ (density), we get:
$d/M = p/RT$
Rearranging to find density: $d = \frac{pM}{RT}$.
Rearranging to find molar mass: $M = \frac{dRT}{p}$.
These equations show that at constant $T$ and $p$, the density of a gas is directly proportional to its molar mass. Also, at constant $T$ and for a specific gas, density is directly proportional to pressure, and molar mass is inversely proportional to density (as expected, $M$ is a fixed property of the gas, but $d$ changes with $p$ and $T$).
Dalton’s Law Of Partial Pressures
John Dalton's Law (1801) applies to mixtures of non-reactive gases. It states that the total pressure exerted by a mixture of non-reactive gases in a container is equal to the sum of the partial pressures of each individual gas.
The partial pressure of a gas in a mixture is the pressure that the gas would exert if it were alone in the same volume and at the same temperature.
For a mixture of gases 1, 2, 3, ... in volume $V$ at temperature $T$:
$p_{\text{Total}} = p_1 + p_2 + p_3 + \dots$ (at constant $T, V$)
where $p_1, p_2, \dots$ are the partial pressures.
Gases collected over water contain water vapor. The pressure of the dry gas can be found by subtracting the vapor pressure of water at that temperature (called aqueous tension) from the total pressure:
$p_{\text{Dry gas}} = p_{\text{Total}} - \text{Aqueous tension}$
Temp/°C | Pressure/bar | Temp/°C | Pressure/bar |
0.0 | 0.0060 | 25.0 | 0.0313 |
5.0 | 0.0087 | 30.0 | 0.0424 |
10.0 | 0.0123 | 35.0 | 0.0563 |
15.0 | 0.0168 | 40.0 | 0.0729 |
20.0 | 0.0230 | 50.0 | 0.1234 |
21.0 | 0.0245 | 60.0 | 0.1992 |
22.0 | 0.0262 | 70.0 | 0.3116 |
23.0 | 0.0281 | 80.0 | 0.4739 |
24.0 | 0.0300 | 90.0 | 0.6967 |
100.0 | 1.0133 |
Partial pressure can also be expressed in terms of mole fraction ($x$). For a gas $i$ in a mixture:
$p_i = x_i p_{\text{Total}}$
where $x_i = n_i / n_{\text{Total}}$, and $n_{\text{Total}}$ is the total number of moles of all gases in the mixture.
Problem 5.4. A neon-dioxygen mixture contains 70.6 g dioxygen and 167.5 g neon. If pressure of the mixture of gases in the cylinder is 25 bar. What is the partial pressure of dioxygen and neon in the mixture ?
Answer:
To find the partial pressure of each gas, we can use the formula $p_i = x_i p_{\text{Total}}$. First, we need to calculate the mole fraction of each gas.
Given mass of dioxygen (O$_2$) = 70.6 g.
Given mass of neon (Ne) = 167.5 g.
Given total pressure $p_{\text{Total}} = 25$ bar.
Step 1: Calculate the number of moles of each gas.
Molar mass of O$_2$ = $2 \times 16.00$ g/mol = 32.00 g/mol.
Moles of O$_2$ ($n_{\text{O}_2}$) = $\frac{\text{Mass of O}_2}{\text{Molar mass of O}_2} = \frac{70.6 \text{ g}}{32.00 \text{ g/mol}} \approx 2.206$ mol.
Molar mass of Ne = 20.18 g/mol (from periodic table). Text uses 20.1 g/mol. Let's use 20.1 g/mol as per text example.
Moles of Ne ($n_{\text{Ne}}$) = $\frac{\text{Mass of Ne}}{\text{Molar mass of Ne}} = \frac{167.5 \text{ g}}{20.1 \text{ g/mol}} \approx 8.333$ mol.
Step 2: Calculate the total number of moles in the mixture.
$n_{\text{Total}} = n_{\text{O}_2} + n_{\text{Ne}} = 2.206 \text{ mol} + 8.333 \text{ mol} = 10.539$ mol.
Step 3: Calculate the mole fraction of each gas.
Mole fraction of O$_2$ ($x_{\text{O}_2}$) = $\frac{n_{\text{O}_2}}{n_{\text{Total}}} = \frac{2.206 \text{ mol}}{10.539 \text{ mol}} \approx 0.2093$. (Text gets 0.21)
Mole fraction of Ne ($x_{\text{Ne}}$) = $\frac{n_{\text{Ne}}}{n_{\text{Total}}} = \frac{8.333 \text{ mol}}{10.539 \text{ mol}} \approx 0.7907$. (Text gets 0.79)
Check: $x_{\text{O}_2} + x_{\text{Ne}} = 0.2093 + 0.7907 = 1$.
Step 4: Calculate the partial pressure of each gas using $p_i = x_i p_{\text{Total}}$.
Partial pressure of O$_2$ ($p_{\text{O}_2}$) = $x_{\text{O}_2} \times p_{\text{Total}} = 0.2093 \times 25 \text{ bar} \approx 5.2325$ bar.
Partial pressure of Ne ($p_{\text{Ne}}$) = $x_{\text{Ne}} \times p_{\text{Total}} = 0.7907 \times 25 \text{ bar} \approx 19.7675$ bar.
Rounding to two decimal places as in text's result:
Partial pressure of O$_2$ $\approx 5.23$ bar.
Partial pressure of Ne $\approx 19.77$ bar.
Check: $p_{\text{O}_2} + p_{\text{Ne}} = 5.23 + 19.77 = 25.00$ bar, which equals $p_{\text{Total}}$.
Using the rounded mole fractions from the text (0.21 and 0.79):
Partial pressure of O$_2$ = $0.21 \times 25 \text{ bar} = 5.25$ bar.
Partial pressure of Ne = $0.79 \times 25 \text{ bar} = 19.75$ bar.
The partial pressure of dioxygen is approximately 5.25 bar, and the partial pressure of neon is approximately 19.75 bar.
Kinetic Energy And Molecular Speeds
Gas molecules are in continuous, random motion, colliding with each other and the container walls. These collisions cause molecules to have a range of different speeds and kinetic energies at any given instant. Thus, we describe molecular speeds using averages.
Different types of average speeds:
- Average speed ($u_{av}$): The arithmetic mean of the speeds of all molecules: $u_{av} = \frac{u_1 + u_2 + \dots + u_n}{n}$
- Root mean square speed ($u_{rms}$): The square root of the average of the squares of the speeds: $u_{rms} = \sqrt{\frac{u_1^2 + u_2^2 + \dots + u_n^2}{n}} = \sqrt{\bar{u^2}}$. This is related to the average kinetic energy.
- Most probable speed ($u_{mp}$): The speed possessed by the largest number of molecules in the gas sample.
The distribution of molecular speeds in a gas sample at a given temperature is described by the Maxwell-Boltzmann distribution of speeds.
At a given temperature, the most probable speed corresponds to the peak of the curve. Higher temperatures shift the distribution towards higher speeds and broaden the curve, indicating a larger fraction of molecules moving at higher speeds. At the same temperature, lighter gases have higher average speeds than heavier gases.
The relationship between the three speeds is: $u_{rms} > u_{av} > u_{mp}$. Their ratio is approximately $1.224 : 1.128 : 1$.
$u_{mp} : u_{av} : u_{rms} :: 1 : 1.128 : 1.224$
The average kinetic energy of gas molecules is directly proportional to the absolute temperature. $\text{KE}_{\text{avg}} = \frac{3}{2} kT$, where $k$ is Boltzmann constant ($R/N_A$). This is a key conclusion from the kinetic theory of gases.
Kinetic Molecular Theory Of Gases
The Kinetic Molecular Theory of Gases is a theoretical model that explains the observed behavior of gases (the gas laws) based on assumptions about the nature and motion of gas particles. It provides a microscopic perspective on gas properties.
Key postulates of the kinetic-molecular theory:
- Gases consist of a very large number of identical particles (atoms or molecules) that are in constant random motion.
- The actual volume occupied by the gas particles is negligible compared to the total volume of the container. Particles are treated as point masses. This explains gas compressibility and why gases fill their container.
- There are no attractive or repulsive forces between gas particles (at ordinary temperature and pressure). This explains why gases expand freely.
- Gas particles move in straight lines and collide with each other and the walls of the container. Collisions with the walls exert pressure.
- Collisions are perfectly elastic. Total kinetic energy is conserved during collisions, although energy can be transferred between particles. This explains why gases don't lose energy and settle down.
- At any given instant, particles have different speeds and kinetic energies. The distribution of speeds remains constant at a constant temperature.
- The average kinetic energy of the gas particles is directly proportional to the absolute temperature. Higher temperature means higher average kinetic energy and faster motion.
These postulates, particularly assumptions 2 and 3, describe an ideal gas. The kinetic theory successfully derives the gas laws, confirming the validity of this model for explaining gas behavior under ideal conditions.
Behaviour Of Real Gases: Deviation From Ideal Gas Behaviour
Real gases do not strictly obey the ideal gas law ($pV=nRT$) under all conditions, particularly at high pressures and low temperatures. Their behavior deviates from the ideal model because the assumptions about negligible particle volume and absence of intermolecular forces are not entirely accurate for real gases.
To observe deviations, we plot $pV$ vs $p$. For an ideal gas, $pV$ is constant at constant $T$, so the plot is a horizontal line. For real gases, the plot deviates significantly.
Some gases (like H$_2$, He) show positive deviation ($pV > nRT$ or $Z > 1$) at most pressures, meaning $pV$ increases with $p$. Other gases (like CO, CH$_4$) show negative deviation ($pV < nRT$ or $Z < 1$) at intermediate pressures (pV decreases with $p$ initially), reaching a minimum, and then show positive deviation at high pressures.
Deviation from ideal behaviour is also seen in $p$ vs $V$ plots. Real gas $p-V$ curves diverge from ideal curves at high pressures, where the actual volume is larger than predicted by ideal gas law.
Reasons for deviation from ideal behavior:
- Intermolecular forces are NOT negligible: At high pressures, molecules are close together, and attractive forces become significant. These forces pull molecules towards each other, reducing their impact on the container walls, resulting in a lower measured pressure than predicted by the ideal gas law. At very low temperatures, thermal energy is low, and attractive forces can cause molecules to stick together, leading to liquefaction.
- Volume of gas molecules is NOT negligible: At high pressures, the total volume occupied by the gas particles themselves becomes a significant fraction of the container volume. The space available for molecules to move in is effectively reduced ($V_{available} = V_{container} - V_{molecule}$). This "excluded volume" makes the actual volume larger than predicted by the ideal gas law if $V$ is taken as the container volume.
Van der Waals Equation: Johannes van der Waals proposed a modified ideal gas equation to account for these deviations, introducing correction terms for pressure and volume:
$(p + \frac{an^2}{V^2})(V - nb) = nRT$
Here, $a$ and $b$ are van der Waals constants specific to each gas. $\frac{an^2}{V^2}$ is the pressure correction term (accounting for intermolecular attractions). $nb$ is the volume correction term (accounting for the volume of the particles). Constant $a$ is a measure of the strength of intermolecular attractive forces. Constant $b$ is related to the effective size of the molecules.
Compressibility Factor (Z): A quantitative measure of deviation from ideal behavior is the compressibility factor, $Z = \frac{pV}{nRT}$.
- For an ideal gas, $Z=1$ at all $p$ and $T$.
- For real gases, $Z$ deviates from 1.
- If $Z < 1$ ($pV < nRT$), the gas is more compressible than ideal; attractive forces are dominant.
- If $Z > 1$ ($pV > nRT$), the gas is less compressible than ideal; repulsive forces or molecular volume effects are dominant.
At very low pressures, $Z \approx 1$ for all gases (ideal behavior). At high pressures, $Z > 1$ for all gases (molecular volume becomes significant, repulsions dominate). At intermediate pressures, most gases have $Z < 1$ (attractions dominate). The deviation depends on the gas and temperature.
The Boyle temperature (or Boyle point) is the temperature at which a real gas obeys the ideal gas law over a significant range of pressure. Above the Boyle temperature, real gases show positive deviations ($Z > 1$). Below the Boyle temperature, they show negative deviations ($Z < 1$) at lower pressures and positive deviations ($Z > 1$) at higher pressures.
Real gases approach ideal behavior at low pressure and high temperature, where molecules are far apart and move fast, making intermolecular forces and molecular volume negligible.
Liquifaction Of Gases
Gases can be converted into liquids by cooling and/or compressing them. The first comprehensive study on the transition between gaseous and liquid states was conducted by Thomas Andrews on carbon dioxide, plotting isotherms ($p$ vs $V$ at constant $T$).
Andrews' observations for CO$_2$:
- At high temperatures (e.g., 50$^\circ$C), the isotherms resemble those of an ideal gas, and increasing pressure simply compresses the gas without liquefaction.
- As temperature decreases, deviations appear. At 30.98$^\circ$C, increasing pressure compresses the gas until a point (E) is reached where liquid first appears at 73 atm pressure. This temperature (30.98$^\circ$C) is the critical temperature ($T_C$) for CO$_2$, the highest temperature at which liquid CO$_2$ can exist. The pressure at this point is the critical pressure ($p_C$) (73 atm). The volume occupied by one mole of gas at $T_C$ and $p_C$ is the critical volume ($V_C$). These are the critical constants. At $T_C$, the liquid and gas phases become indistinguishable.
- Below $T_C$ (e.g., 21.5$^\circ$C), compression follows the gas behavior (A to B). At point B, liquid appears, and gas starts to condense. Pressure remains constant during condensation (B to C) as more gas turns into liquid. At point C, all gas is liquefied. Further compression steeply increases pressure because liquids are nearly incompressible (C to D). The region under the dome-shaped curve represents the coexistence of liquid and gas in equilibrium.
For liquefaction to occur, a gas must be cooled to a temperature at or below its critical temperature ($T_C$). Compression below $T_C$ then brings molecules close enough for intermolecular forces to become dominant, leading to the formation of the liquid phase.
There is a continuity between the gaseous and liquid states. A substance can change from gas to liquid or liquid to gas without passing through a two-phase region, especially at or above the critical temperature. The term fluid is used for either a liquid or a gas. A liquid is essentially a dense fluid. The distinction between liquid and gas exists only below $T_C$ where the two phases can coexist in equilibrium with a visible boundary.
A gas below its critical temperature is often called a vapor. For example, CO$_2$ gas below 30.98$^\circ$C is called CO$_2$ vapor.
Critical constants for some substances (Table 5.4):
Substance | Critical Temperature/K | Critical Pressure/bar | Critical Volume/dm$^3$ mol$^{-1}$ |
H$_2$ | 33.2 | 13.6 | 0.065 |
He | 5.3 | 2.3 | 0.057 |
N$_2$ | 126.0 | 33.9 | 0.090 |
O$_2$ | 154.3 | 50.8 | 0.074 |
CO$_2$ | 304.1 | 73.9 | 0.095 |
H$_2$O | 647.1 | 220.6 | 0.056 |
NH$_3$ | 405.5 | 113.0 | 0.072 |
Gases with higher critical temperatures have stronger intermolecular forces, requiring less cooling for liquefaction.
Problem 5.5. Gases possess characteristic critical temperature which depends upon the magnitude of intermolecular forces between the gas particles. Critical temperatures of ammonia and carbon dioxide are 405.5 K and 304.10 K respectively. Which of these gases will liquify first when you start cooling from 500 K to their critical temperature ?
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Liquefaction of a gas by cooling begins when its temperature reaches or falls below its critical temperature ($T_C$).
Given starting temperature: 500 K.
Critical temperature of ammonia (NH$_3$): $T_C(\text{NH}_3) = 405.5$ K.
Critical temperature of carbon dioxide (CO$_2$): $T_C(\text{CO}_2) = 304.10$ K.
We are cooling the gases from 500 K. The gas that reaches its critical temperature first will be the first one that can potentially liquefy upon cooling (with sufficient pressure).
As we cool from 500 K:
- The temperature will pass 405.5 K first (for NH$_3$).
- Then it will continue to cool and later pass 304.10 K (for CO$_2$).
Ammonia reaches its critical temperature (405.5 K) before carbon dioxide reaches its critical temperature (304.10 K).
Therefore, Ammonia will liquefy first upon cooling from 500 K (provided sufficient pressure is applied at or below 405.5 K).
The higher critical temperature of NH$_3$ compared to CO$_2$ also indicates that NH$_3$ has stronger intermolecular forces.
Liquid State
In the liquid state, intermolecular forces are strong enough to hold molecules close together, giving liquids a definite volume, but weak enough to allow molecules to move past one another, allowing liquids to flow and take the shape of their container. Liquids are much denser than gases due to the close proximity of molecules and have very little empty space.
Some characteristic physical properties of liquids stemming from intermolecular attractions include vapour pressure, surface tension, and viscosity.
Vapour Pressure
When a liquid is placed in a closed container, some molecules with sufficient kinetic energy escape from the liquid surface into the space above as vapour (evaporation). The vapour particles collide with the container walls, exerting pressure (vapour pressure). As evaporation proceeds, the concentration of vapour molecules increases, and some vapour molecules return to the liquid state (condensation).
Eventually, a dynamic equilibrium is established where the rate of evaporation equals the rate of condensation. The pressure exerted by the vapour in equilibrium with the liquid at a given temperature is called the equilibrium vapour pressure or saturated vapour pressure. Vapour pressure increases with temperature because higher thermal energy leads to more molecules having enough energy to escape the liquid phase. Vapour pressure values are temperature-dependent.
Boiling Point: When a liquid is heated in an open container, it boils when its vapour pressure becomes equal to the external atmospheric pressure. The temperature at which this occurs is the boiling point at that external pressure. The normal boiling point is the boiling point at 1 atm pressure (101.325 kPa). The standard boiling point is the boiling point at 1 bar pressure (100 kPa), which is slightly lower than the normal boiling point.
At high altitudes, atmospheric pressure is lower, so liquids boil at lower temperatures. Pressure cookers are used at high altitudes to increase pressure and raise the boiling point of water, allowing food to cook faster.
In a closed vessel, heating a liquid increases vapour pressure. As temperature approaches the critical temperature, the densities of the liquid and vapour phases become equal, and the boundary between them disappears. This is the critical point.
A gas below its critical temperature is called a vapour and can be liquefied by applying pressure. Above the critical temperature, it is called a gas and cannot be liquefied by pressure alone.
Surface Tension
Liquids exhibit surface tension, a property that arises from the imbalance of intermolecular forces experienced by molecules at the surface compared to those in the bulk of the liquid. A molecule in the interior of a liquid is surrounded by other molecules and experiences attractive forces equally in all directions, resulting in no net force. However, a molecule at the liquid surface experiences attractive forces primarily from molecules below and to the sides, with no attractive forces from above (assuming no air or gas molecules interacting).
This imbalance results in a net inward (downward) attractive force on surface molecules, pulling them towards the bulk of the liquid. To increase the surface area of a liquid, molecules from the bulk must be moved to the surface, working against these inward forces, which requires energy. This energy required to increase the surface area by one unit is called surface energy (units: J m$^{-2}$).
Surface tension ($\gamma$) is defined as the force acting per unit length perpendicular to a line drawn on the liquid surface (units: N m$^{-1}$). Surface tension is numerically equal to surface energy. Liquids tend to minimize their surface area to minimise surface energy, which is why small liquid drops in the absence of gravity are spherical (a sphere has the minimum surface area for a given volume).
Surface tension explains phenomena like capillary action (liquid rising or falling in narrow tubes), the spherical shape of droplets, and the ability of some insects to walk on water. Stronger intermolecular forces lead to higher surface tension. Increasing temperature decreases surface tension because increased kinetic energy weakens the effectiveness of intermolecular attractions.
Viscosity
Viscosity is a measure of a liquid's resistance to flow. It results from the internal friction between different layers of liquid as they move past each other. Stronger intermolecular forces lead to greater resistance to flow and higher viscosity.
When a liquid flows over a surface (e.g., down a pipe), the layer of liquid in contact with the surface is stationary. Layers further away move at increasing velocities. This type of flow, with a velocity gradient between layers, is called laminar flow.
A force ($F$) is needed to maintain laminar flow, proportional to the area of contact ($A$) between the layers and the velocity gradient ($\frac{du}{dz}$, where $du$ is the change in velocity between layers separated by distance $dz$).
$F \propto A \frac{du}{dz}$
$F = \eta A \frac{du}{dz}$
where $\eta$ (eta) is the proportionality constant called the coefficient of viscosity. It quantifies the viscosity of the liquid. A higher value of $\eta$ means higher viscosity and slower flow.
SI unit of viscosity coefficient is Pascal-second (Pa s) or N s m$^{-2}$ or kg m$^{-1}$ s$^{-1}$. CGS unit is poise (P): 1 poise = $1 \text{ g cm}^{-1} \text{ s}^{-1} = 0.1 \text{ kg m}^{-1} \text{ s}^{-1}$.
Viscosity of liquids generally decreases as temperature increases. Higher temperature means greater thermal energy, which helps molecules overcome intermolecular forces and slip past each other more easily.
Exercises
Question 5.1 What will be the minimum pressure required to compress 500 $dm^3$ of air at 1 bar to 200 $dm^3$ at $30^\circ C$?
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Question 5.2 A vessel of 120 mL capacity contains a certain amount of gas at $35 ^\circ C$ and 1.2 bar pressure. The gas is transferred to another vessel of volume 180 mL at $35 ^\circ C$. What would be its pressure?
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Question 5.3 Using the equation of state $pV=nRT$; show that at a given temperature density of a gas is proportional to gas pressure p.
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Question 5.4 At $0^\circ C$, the density of a certain oxide of a gas at 2 bar is same as that of dinitrogen at 5 bar. What is the molecular mass of the oxide?
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Question 5.5 Pressure of 1 g of an ideal gas A at $27 ^\circ C$ is found to be 2 bar. When 2 g of another ideal gas B is introduced in the same flask at same temperature the pressure becomes 3 bar. Find a relationship between their molecular masses.
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Question 5.6 The drain cleaner, Drainex contains small bits of aluminum which react with caustic soda to produce dihydrogen. What volume of dihydrogen at $20 ^\circ C$ and one bar will be released when 0.15g of aluminum reacts?
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Question 5.7 What will be the pressure exerted by a mixture of 3.2 g of methane and 4.4 g of carbon dioxide contained in a 9 $dm^3$ flask at $27 ^\circ C$ ?
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Question 5.8 What will be the pressure of the gaseous mixture when 0.5 L of $H_2$ at 0.8 bar and 2.0 L of dioxygen at 0.7 bar are introduced in a 1L vessel at $27^\circ C$?
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Question 5.9 Density of a gas is found to be 5.46 $g/dm^3$ at $27 ^\circ C$ at 2 bar pressure. What will be its density at STP?
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Question 5.10 34.05 mL of phosphorus vapour weighs 0.0625 g at $546 ^\circ C$ and 0.1 bar pressure. What is the molar mass of phosphorus?
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Question 5.11 A student forgot to add the reaction mixture to the round bottomed flask at $27 ^\circ C$ but instead he/she placed the flask on the flame. After a lapse of time, he realized his mistake, and using a pyrometer he found the temperature of the flask was $477 ^\circ C$. What fraction of air would have been expelled out?
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Question 5.12 Calculate the temperature of 4.0 mol of a gas occupying 5 $dm^3$ at 3.32 bar. ($R = 0.083 \ bar \ dm^3 \ K^{–1} \ mol^{–1}$).
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Question 5.13 Calculate the total number of electrons present in 1.4 g of dinitrogen gas.
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Question 5.14 How much time would it take to distribute one Avogadro number of wheat grains, if $10^{10}$ grains are distributed each second ?
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Question 5.15 Calculate the total pressure in a mixture of 8 g of dioxygen and 4 g of dihydrogen confined in a vessel of 1 $dm^3$ at $27^\circ C$. R = $0.083 \ bar \ dm^3 \ K^{–1} \ mol^{–1}$.
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Question 5.16 Pay load is defined as the difference between the mass of displaced air and the mass of the balloon. Calculate the pay load when a balloon of radius 10 m, mass 100 kg is filled with helium at 1.66 bar at $27^\circ C$. (Density of air = $1.2 \ kg \ m^{–3}$ and $R = 0.083 \ bar \ dm^3 \ K^{–1} \ mol^{–1}$).
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Question 5.17 Calculate the volume occupied by 8.8 g of $CO_2$ at $31.1^\circ C$ and 1 bar pressure. $R = 0.083 \ bar \ L \ K^{–1} \ mol^{–1}$.
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Question 5.18 2.9 g of a gas at $95 ^\circ C$ occupied the same volume as 0.184 g of dihydrogen at $17 ^\circ C$, at the same pressure. What is the molar mass of the gas?
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Question 5.19 A mixture of dihydrogen and dioxygen at one bar pressure contains 20% by weight of dihydrogen. Calculate the partial pressure of dihydrogen.
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Question 5.20 What would be the SI unit for the quantity $pV^2T^2/n$ ?
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Question 5.21 In terms of Charles’ law explain why $–273 ^\circ C$ is the lowest possible temperature.
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Question 5.22 Critical temperature for carbon dioxide and methane are $31.1 ^\circ C$ and $–81.9 ^\circ C$ respectively. Which of these has stronger intermolecular forces and why?
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Question 5.23 Explain the physical significance of van der Waals parameters.
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