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Gaseous and Liquid States



States Of Matter (Gaseous & Liquid Properties)

As discussed previously, matter exists primarily in three physical states: solid, liquid, and gas. These states are distinguished by the arrangement, movement, and interaction of the constituent particles (atoms or molecules). While solids have definite shape and volume, liquids have definite volume but no definite shape, and gases have neither definite shape nor definite volume.

Understanding the properties of the gaseous and liquid states is crucial in chemistry as many chemical reactions and physical processes involve substances in these states.


The Gaseous State

The gaseous state is the simplest state of matter. Gases are highly compressible and expand to fill the entire volume of the container they are placed in. They exert pressure equally in all directions.

The properties of gases are primarily determined by the large distances between particles and the weak intermolecular forces compared to liquids and solids. The particles are in constant, random, high-speed motion.

Observable properties of gases include volume, pressure, temperature, and amount (number of moles).




The Gas Laws

The relationships between the pressure, volume, temperature, and amount of a gas were discovered through experimental observations over centuries. These relationships are summarised in the gas laws.


Boyle’s Law (Pressure - Volume Relationship)

Boyle's Law describes the inverse relationship between the pressure and volume of a fixed amount of gas at a constant temperature. It was given by Robert Boyle in 1662.

Statement: At constant temperature, the volume of a fixed mass of a gas is inversely proportional to its pressure.

Mathematically, this can be expressed as:

$ V \propto \frac{1}{P} \quad (\text{at constant } n, T) $

This proportionality can be written as an equation by introducing a constant, $k_1$:

$ V = \frac{k_1}{P} \quad \text{or} \quad PV = k_1 $

This means that for a given mass of gas at a constant temperature, the product of its pressure and volume is constant. So, if a gas at pressure $P_1$ and volume $V_1$ undergoes a change to pressure $P_2$ and volume $V_2$ at the same temperature and amount, then:

$ P_1V_1 = P_2V_2 $

Graphically, the relationship between P and V at constant temperature is a hyperbola. A plot of P vs 1/V is a straight line passing through the origin.

Graphs illustrating Boyle's Law (P vs V and P vs 1/V)

Example 1. A gas occupies a volume of 10 L at a pressure of 1 atm. What volume will it occupy if the pressure is increased to 2 atm at the same temperature?

Answer:

Given: $V_1 = 10$ L, $P_1 = 1$ atm, $P_2 = 2$ atm.

Temperature and amount of gas are constant. Using Boyle's Law: $P_1V_1 = P_2V_2$

$ (1 \text{ atm}) \times (10 \text{ L}) = (2 \text{ atm}) \times V_2 $

$ V_2 = \frac{1 \text{ atm} \times 10 \text{ L}}{2 \text{ atm}} = 5 \text{ L} $

Answer: The gas will occupy a volume of 5 L.


Charles’ Law (Temperature - Volume Relationship)

Charles' Law describes the direct relationship between the volume and absolute temperature of a fixed amount of gas at constant pressure. It was formulated by Jacques Charles in 1787 and independently by Joseph Louis Gay-Lussac in 1802.

Statement: At constant pressure, the volume of a fixed mass of a gas is directly proportional to its absolute temperature.

Mathematically:

$ V \propto T \quad (\text{at constant } n, P) $

This can be written as:

$ V = k_2 T \quad \text{or} \quad \frac{V}{T} = k_2 $

So, for a gas undergoing a change from $(V_1, T_1)$ to $(V_2, T_2)$ at constant pressure and amount:

$ \frac{V_1}{T_1} = \frac{V_2}{T_2} $

Note: Temperature must be in Kelvin (K).

A plot of V vs T (in Kelvin) at constant pressure is a straight line passing through the origin. If the temperature is plotted in Celsius ($^\circ$C), the line intercepts the temperature axis at -273.15$^\circ$C. This temperature is called absolute zero (0 K), the lowest possible temperature.

Graphs illustrating Charles' Law (V vs T in K and V vs T in Celsius)

Example 2. A gas occupies a volume of 500 mL at 27$^\circ$C. What volume will it occupy at 127$^\circ$C if the pressure remains constant?

Answer:

Given: $V_1 = 500$ mL, $T_1 = 27^\circ$C, $T_2 = 127^\circ$C.

Convert temperatures to Kelvin:

$T_1 = 27 + 273.15 = 300.15$ K (approx 300 K for simplicity)

$T_2 = 127 + 273.15 = 400.15$ K (approx 400 K)

Using Charles' Law: $\frac{V_1}{T_1} = \frac{V_2}{T_2}$

$ \frac{500 \text{ mL}}{300 \text{ K}} = \frac{V_2}{400 \text{ K}} $

$ V_2 = \frac{500 \text{ mL} \times 400 \text{ K}}{300 \text{ K}} = \frac{200000}{300} \text{ mL} \approx 666.67 \text{ mL} $

Answer: The gas will occupy approximately 666.67 mL.


Gay Lussac’s Law (Pressure- Temperature Relationship)

Gay-Lussac's Law describes the direct relationship between the pressure and absolute temperature of a fixed amount of gas at constant volume. It was stated by Joseph Louis Gay-Lussac in 1802.

Statement: At constant volume, the pressure of a fixed mass of a gas is directly proportional to its absolute temperature.

Mathematically:

$ P \propto T \quad (\text{at constant } n, V) $

This can be written as:

$ P = k_3 T \quad \text{or} \quad \frac{P}{T} = k_3 $

So, for a gas undergoing a change from $(P_1, T_1)$ to $(P_2, T_2)$ at constant volume and amount:

$ \frac{P_1}{T_1} = \frac{P_2}{T_2} $

Note: Temperature must be in Kelvin (K).

A plot of P vs T (in Kelvin) at constant volume is a straight line passing through the origin.

Graph illustrating Gay-Lussac's Law (P vs T)

Example 3. A gas in a sealed container has a pressure of 1.5 atm at 27$^\circ$C. What will be the pressure at 100$^\circ$C if the volume remains constant?

Answer:

Given: $P_1 = 1.5$ atm, $T_1 = 27^\circ$C, $T_2 = 100^\circ$C.

Convert temperatures to Kelvin:

$T_1 = 27 + 273.15 = 300.15$ K (approx 300 K)

$T_2 = 100 + 273.15 = 373.15$ K (approx 373 K)

Using Gay-Lussac's Law: $\frac{P_1}{T_1} = \frac{P_2}{T_2}$

$ \frac{1.5 \text{ atm}}{300 \text{ K}} = \frac{P_2}{373 \text{ K}} $

$ P_2 = \frac{1.5 \text{ atm} \times 373 \text{ K}}{300 \text{ K}} = \frac{559.5}{300} \text{ atm} \approx 1.865 \text{ atm} $

Answer: The pressure will be approximately 1.865 atm.


Avogadro Law (Volume - Amount Relationship)

Avogadro's Law describes the direct relationship between the volume and the amount (number of moles) of a gas at constant temperature and pressure. It was proposed by Amedeo Avogadro in 1811.

Statement: At the same temperature and pressure, equal volumes of all gases contain an equal number of molecules (or moles).

Mathematically:

$ V \propto n \quad (\text{at constant } T, P) $

This can be written as:

$ V = k_4 n \quad \text{or} \quad \frac{V}{n} = k_4 $

So, for two gases A and B at the same temperature and pressure:

$ \frac{V_A}{n_A} = \frac{V_B}{n_B} $

Avogadro's law implies that the molar volume (volume occupied by one mole of any gas) is constant at a given temperature and pressure.


Example 4. If 2 moles of a gas occupy a volume of 44.8 L at STP, what volume will 0.5 moles of the same gas occupy at STP?

Answer:

Given: $n_1 = 2$ mol, $V_1 = 44.8$ L, $n_2 = 0.5$ mol.

Temperature and pressure are constant (STP). Using Avogadro's Law: $\frac{V_1}{n_1} = \frac{V_2}{n_2}$

$ \frac{44.8 \text{ L}}{2 \text{ mol}} = \frac{V_2}{0.5 \text{ mol}} $

$ V_2 = \frac{44.8 \text{ L} \times 0.5 \text{ mol}}{2 \text{ mol}} = \frac{22.4}{2} \text{ L} = 11.2 \text{ L} $

Answer: 0.5 moles of the gas will occupy 11.2 L at STP.




Ideal Gas Equation

The ideal gas equation is a single equation that combines Boyle's Law, Charles' Law, and Avogadro's Law. It describes the behaviour of an ideal gas, which is a hypothetical gas that perfectly follows the gas laws.

From the individual gas laws, we have:

Combining these proportionalities, we get:

$ V \propto \frac{nT}{P} $

Introducing a constant of proportionality, R, known as the Ideal Gas Constant or Universal Gas Constant, we get the ideal gas equation:

$ PV = nRT $

Where:

The value of R depends on the units used for pressure and volume. Some common values of R are:

The ideal gas equation is useful for calculating one of the variables (P, V, n, or T) if the other three are known, or for describing the state changes of a gas ($P_1V_1/T_1 = P_2V_2/T_2$ when n is constant).


Example 5. Calculate the volume occupied by 0.1 moles of an ideal gas at 27$^\circ$C and 1 atm pressure.

Answer:

Given: $n = 0.1$ mol, $T = 27^\circ$C, $P = 1$ atm.

Convert temperature to Kelvin: $T = 27 + 273.15 = 300.15$ K (approx 300 K)

Using the ideal gas equation $PV = nRT$. Since P is in atm and we want V in L, we use R = 0.0821 L atm / (mol K).

$ (1 \text{ atm}) \times V = (0.1 \text{ mol}) \times (0.0821 \text{ L atm / (mol K)}) \times (300 \text{ K}) $

$ V = \frac{0.1 \times 0.0821 \times 300}{1} \text{ L} $

$ V = 2.463 \text{ L} $

Answer: The volume occupied is approximately 2.463 L.


Density And Molar Mass Of A Gaseous Substance

The ideal gas equation ($PV = nRT$) can be used to relate the density (d) and molar mass (M) of a gas to its pressure and temperature.

We know that the number of moles ($n$) is given by:

$ n = \frac{\text{mass (m)}}{\text{molar mass (M)}} $

Substitute this into the ideal gas equation:

$ PV = \left(\frac{m}{M}\right) RT $

Rearrange the equation to group mass and volume:

$ P = \left(\frac{m}{V}\right) \frac{RT}{M} $

Since density ($d$) is mass per unit volume ($d = m/V$), we can write:

$ P = d \frac{RT}{M} $

This equation can be rearranged to find density or molar mass:

Density:

$ d = \frac{PM}{RT} $

Molar Mass:

$ M = \frac{dRT}{P} \quad \text{or} \quad M = \frac{mRT}{PV} $

These formulas show that the density of a gas is directly proportional to its pressure and molar mass, and inversely proportional to its temperature.


Example 6. Calculate the density of oxygen gas (O$_2$) at 1 atm pressure and 27$^\circ$C.

(Molar mass of O$_2$ = 32 g/mol, R = 0.0821 L atm / (mol K))

Answer:

Given: $P = 1$ atm, $T = 27^\circ$C, $M = 32$ g/mol.

Convert temperature to Kelvin: $T = 27 + 273.15 = 300.15$ K (approx 300 K)

Using the formula for density: $d = \frac{PM}{RT}$

$ d = \frac{(1 \text{ atm}) \times (32 \text{ g/mol})}{(0.0821 \text{ L atm / (mol K)}) \times (300 \text{ K})} $

$ d = \frac{32}{0.0821 \times 300} \text{ g/L} $

$ d = \frac{32}{24.63} \text{ g/L} \approx 1.30 \text{ g/L} $

Answer: The density of oxygen gas is approximately 1.30 g/L.


Dalton’s Law Of Partial Pressures

Dalton's Law of Partial Pressures describes the behaviour of a mixture of non-reacting gases.

Statement: The total pressure exerted by a mixture of non-reacting gases in a given volume and at a constant temperature is equal to the sum of the partial pressures of the individual gases.

The partial pressure of a gas in a mixture is the pressure that the gas would exert if it alone occupied the same volume at the same temperature.

Mathematically, for a mixture of gases A, B, C, etc., the total pressure ($P_{total}$) is:

$ P_{total} = P_A + P_B + P_C + \dots $

Where $P_A, P_B, P_C$ are the partial pressures of gases A, B, C, respectively.

If each gas in the mixture behaves ideally, its partial pressure can be calculated using the ideal gas equation for that specific gas:

$P_A = \frac{n_A RT}{V}$, $P_B = \frac{n_B RT}{V}$, etc.

So, the total pressure is:

$ P_{total} = \frac{n_A RT}{V} + \frac{n_B RT}{V} + \frac{n_C RT}{V} + \dots = \frac{(n_A + n_B + n_C + \dots) RT}{V} = \frac{n_{total} RT}{V} $

Where $n_{total}$ is the total number of moles in the mixture.

The partial pressure of a gas is also related to the total pressure by its mole fraction ($x_i$), which is the ratio of the number of moles of that gas to the total number of moles in the mixture ($x_i = n_i / n_{total}$).

$ P_i = x_i \times P_{total} $

Dalton's law is particularly useful when gases are collected over water, as the collected gas will be a mixture of the dry gas and water vapour. The total pressure is the sum of the pressure of the dry gas and the pressure of water vapour (aqueous tension) at that temperature.

$ P_{total} = P_{dry\ gas} + P_{water\ vapour} $


Example 7. A mixture of 0.5 moles of Nitrogen gas and 0.3 moles of Oxygen gas is placed in a 10 L container at 27$^\circ$C. Calculate the total pressure and the partial pressures of each gas.

(R = 0.0821 L atm / (mol K))

Answer:

Given: $n_{N_2} = 0.5$ mol, $n_{O_2} = 0.3$ mol, $V = 10$ L, $T = 27^\circ$C.

Convert temperature to Kelvin: $T = 27 + 273.15 = 300.15$ K (approx 300 K)

Total number of moles, $n_{total} = n_{N_2} + n_{O_2} = 0.5 \text{ mol} + 0.3 \text{ mol} = 0.8 \text{ mol}$.

Using the ideal gas equation for the total pressure: $P_{total}V = n_{total}RT$

$ P_{total} = \frac{n_{total}RT}{V} = \frac{(0.8 \text{ mol}) \times (0.0821 \text{ L atm / (mol K)}) \times (300 \text{ K})}{10 \text{ L}} $

$ P_{total} = \frac{0.8 \times 0.0821 \times 300}{10} \text{ atm} = \frac{19.704}{10} \text{ atm} = 1.9704 \text{ atm} $

To find partial pressures using mole fractions:

Mole fraction of Nitrogen, $x_{N_2} = \frac{n_{N_2}}{n_{total}} = \frac{0.5 \text{ mol}}{0.8 \text{ mol}} = 0.625$

Mole fraction of Oxygen, $x_{O_2} = \frac{n_{O_2}}{n_{total}} = \frac{0.3 \text{ mol}}{0.8 \text{ mol}} = 0.375$

Partial pressure of Nitrogen, $P_{N_2} = x_{N_2} \times P_{total} = 0.625 \times 1.9704 \text{ atm} \approx 1.2315 \text{ atm}$

Partial pressure of Oxygen, $P_{O_2} = x_{O_2} \times P_{total} = 0.375 \times 1.9704 \text{ atm} \approx 0.7389 \text{ atm}$

Check: $P_{N_2} + P_{O_2} = 1.2315 + 0.7389 = 1.9704 \text{ atm} = P_{total}$.

Answer: Total pressure is approximately 1.9704 atm. Partial pressure of Nitrogen is approximately 1.2315 atm, and partial pressure of Oxygen is approximately 0.7389 atm.




Kinetic Energy And Molecular Speeds

The particles of a gas are in continuous, random motion. This motion gives them kinetic energy. At any given temperature, the particles in a sample of gas have a distribution of speeds and thus kinetic energies. However, the average kinetic energy of the gas particles is directly proportional to the absolute temperature of the gas.

Average Kinetic Energy:

For one mole of an ideal gas, the average kinetic energy (KE$_{avg}$) is given by:

$ \text{KE}_{avg} = \frac{3}{2} RT $

For a single molecule, the average kinetic energy is:

$ \text{KE}_{avg} = \frac{3}{2} kT $

Where $k$ is the Boltzmann constant ($k = R/N_A$, where $N_A$ is Avogadro's number). $k = 1.38 \times 10^{-23}$ J/K.

This relationship ($KE_{avg} \propto T$) is fundamental to understanding the effect of temperature on gas behaviour. Higher the temperature, greater the average kinetic energy and speed of the gas particles.

Since gas particles move at different speeds, we often refer to different types of average speeds:

  1. Most Probable Speed ($u_{mp}$): The speed possessed by the largest fraction of gas molecules at a given temperature.
  2. Average Speed ($\bar{u}$ or $u_{avg}$): The arithmetic mean of the speeds of all the molecules.
  3. Root Mean Square Speed ($u_{rms}$): The square root of the average of the squares of the speeds of all the molecules. This is related to the kinetic energy.

The formulas for these speeds for a gas with molar mass M (in kg/mol) at temperature T (in K) are:

$ u_{mp} = \sqrt{\frac{2RT}{M}} $

$ \bar{u} = \sqrt{\frac{8RT}{\pi M}} $

$ u_{rms} = \sqrt{\frac{3RT}{M}} $

Note the order of these speeds: $u_{mp} < \bar{u} < u_{rms}$ for any given gas at a specific temperature. The ratios are $u_{mp} : \bar{u} : u_{rms} = \sqrt{2} : \sqrt{8/\pi} : \sqrt{3} \approx 1.414 : 1.596 : 1.732$.

These speeds are also inversely proportional to the square root of the molar mass ($u \propto \sqrt{T/M}$). At the same temperature, lighter gases move faster than heavier gases.

The distribution of molecular speeds at different temperatures is shown by the Maxwell-Boltzmann distribution curve.

Maxwell-Boltzmann distribution curve showing molecular speeds at different temperatures

As temperature increases, the distribution broadens, shifts to higher speeds, and the peak (most probable speed) becomes lower, indicating a wider range of speeds and a higher average kinetic energy.




Kinetic Molecular Theory Of Gases

The Kinetic Molecular Theory (KMT) is a model that explains the macroscopic properties of gases (like pressure, volume, and temperature) in terms of the behaviour of their microscopic particles (atoms or molecules). It provides a theoretical basis for the gas laws and the ideal gas equation.

The key postulates of the Kinetic Molecular Theory are:

  1. Gases consist of a large number of identical, tiny particles (atoms or molecules) that are in constant, random motion. The motion is linear and in random directions.
  2. The volume occupied by the gas particles themselves is negligible compared to the total volume of the container. The particles are considered point masses, having mass but negligible volume. (This postulate holds well at low pressures where the gas particles are far apart).
  3. There are no significant forces of attraction or repulsion between gas particles. The particles are assumed to be non-interacting except during collisions. (This holds well at high temperatures and low pressures where particles are far apart).
  4. The collisions between gas particles and between the particles and the walls of the container are perfectly elastic. Elastic collisions mean that there is no net loss of kinetic energy during collisions. Energy can be transferred between particles, but the total kinetic energy of the system remains constant at a given temperature.
  5. The average kinetic energy of the gas particles is directly proportional to the absolute temperature of the gas. $KE_{avg} \propto T$. At a given temperature, the average kinetic energy is the same for all gases.

How KMT explains the Gas Laws:

KMT provides a microscopic explanation for the macroscopic behaviour described by the gas laws.




Behaviour Of Real Gases: Deviation From Ideal Gas Behaviour

The ideal gas equation ($PV = nRT$) and the Kinetic Molecular Theory provide a good description of gas behaviour under certain conditions. However, they are based on assumptions that are not perfectly true for real gases.

Real gases deviate from ideal behaviour, particularly at high pressures and low temperatures.

Reasons for Deviation: Real gases deviate because the postulates of KMT regarding negligible volume of particles and absence of intermolecular forces are not entirely valid for real gases.

  1. Finite Volume of Molecules: At high pressures, the gas particles are forced closer together. The volume occupied by the particles themselves becomes significant compared to the total volume of the container. The effective volume available for the particles to move is less than the container volume ($V - nb$, where $nb$ is the volume correction term). This means the real volume is effectively larger than the ideal volume predicted by the ideal gas equation at high pressures.
  2. Existence of Intermolecular Forces: At low temperatures and high pressures, the gas particles are closer together, and the weak attractive intermolecular forces (like van der Waals forces) become significant. These attractive forces reduce the speed with which particles hit the container walls, leading to a lower pressure than expected for an ideal gas. The observed pressure is effectively less than the ideal pressure ($P + \frac{an^2}{V^2}$, where $\frac{an^2}{V^2}$ is the pressure correction term).

Compressibility Factor (Z): Deviation from ideal behaviour is often measured by the compressibility factor (Z), defined as:

$ Z = \frac{PV_{real}}{nRT} $

For an ideal gas, $Z = 1$ under all conditions.

For real gases:

Graph of Compressibility Factor (Z) vs Pressure for various real gases

The van der Waals equation is a modified ideal gas equation that attempts to account for the finite volume of particles and intermolecular forces in real gases:

$ \left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT $

Where 'a' and 'b' are van der Waals constants that are specific to each gas. 'a' accounts for intermolecular forces, and 'b' accounts for the volume occupied by the gas molecules.

Real gases behave most ideally at low pressures and high temperatures.




Liquifaction Of Gases

Liquefaction of gases is the process of converting a gas into a liquid. This is possible because the attractive forces between gas molecules become significant at high pressures and low temperatures, causing them to condense into a liquid.

The ability to liquefy a gas is closely related to its critical temperature and critical pressure.

A gas can be liquefied only when its temperature is below its critical temperature. At or below $T_c$, increasing the pressure can bring the molecules close enough for the attractive forces to dominate and cause liquefaction.

Andrew's Experiments on CO$_2$: Thomas Andrew (1869) studied the effect of pressure on the volume of CO$_2$ gas at different temperatures. His experiments were crucial in understanding the concept of critical temperature and liquefaction.

He plotted pressure vs volume isotherms for CO$_2$ at various temperatures:

Andrew's isotherms for CO2

Observations from Andrew's isotherms:

Below $T_c$, increasing pressure results in liquefaction. Above $T_c$, the substance remains as a gas (or more accurately, a supercritical fluid above $P_c$).

Common methods for liquefying gases include cooling the gas below its critical temperature and then applying pressure (e.g., Linde's process, Claude's process).

Applications: Storage and transport of gases (like LPG, O$_2$, N$_2$) in liquid form, refrigeration, cryogenics.




Liquid State

The liquid state is an intermediate state between the solid and gaseous states. Liquids possess some properties of both solids (definite volume) and gases (no definite shape, fluidity). The particles in a liquid are close together (like in solids) but are free to move past one another (like in gases, but with less freedom). The intermolecular forces are stronger than in gases but weaker than in solids.

The properties of liquids, such as vapour pressure, surface tension, and viscosity, arise from the intermolecular forces and the ability of particles to move while remaining relatively close.


Vapour Pressure

In a closed container, a liquid is in equilibrium with its vapour. Some particles from the liquid surface with sufficient kinetic energy escape into the space above the liquid as vapour (evaporation). Simultaneously, particles from the vapour collide with the liquid surface and return to the liquid state (condensation).

Eventually, the rate of evaporation becomes equal to the rate of condensation, and a state of dynamic equilibrium is reached.

Vapour Pressure: The pressure exerted by the vapour in equilibrium with its liquid at a given temperature is called its vapour pressure.

Equilibrium between liquid and vapour in a closed container

Factors affecting vapour pressure:

  1. Temperature: Vapour pressure increases with increasing temperature. At higher temperatures, more particles have enough kinetic energy to escape from the liquid surface into the vapour phase, leading to a higher concentration of vapour and thus higher pressure.
  2. Nature of the Liquid: Liquids with weaker intermolecular forces evaporate more easily and have higher vapour pressures at a given temperature compared to liquids with stronger intermolecular forces. Liquids that evaporate easily are called volatile liquids (e.g., petrol, acetone), while those that do not are non-volatile liquids (e.g., water, honey).

Vapour pressure is independent of the amount of liquid or the surface area, as long as both liquid and vapour phases are present in equilibrium.

Boiling Point revisited: The boiling point of a liquid is the temperature at which its vapour pressure becomes equal to the external atmospheric pressure. At this point, the liquid starts boiling, and vaporisation occurs throughout the bulk of the liquid.

If the external pressure is lower, the liquid boils at a lower temperature (e.g., cooking at high altitudes). If the external pressure is higher, the liquid boils at a higher temperature (e.g., in a pressure cooker).

The normal boiling point is the boiling point at standard atmospheric pressure (1 atm or 101.325 kPa).


Surface Tension

Particles within the bulk of a liquid are surrounded by other particles on all sides, experiencing attractive intermolecular forces equally in all directions. However, particles at the surface of the liquid are not surrounded by particles from above. They experience a net inward attractive force from the particles in the bulk of the liquid.

Illustration of forces on liquid molecules in the bulk and at the surface

This net inward force pulls the surface molecules inwards, causing the liquid surface to behave like a stretched membrane. This phenomenon is called surface tension.

Surface Tension ($\gamma$ or $\sigma$): It is defined as the force acting per unit length perpendicular to a line drawn on the surface of a liquid, or as the surface energy per unit area.

$ \text{Surface Tension} = \frac{\text{Force}}{\text{Length}} \quad (\text{Unit: N/m}) $

Or as Surface Energy:

$ \text{Surface Energy} = \frac{\text{Work done}}{\text{Area change}} \quad (\text{Unit: J/m}^2) $

(Note: 1 J/m$^2$ = 1 N m / m$^2$ = 1 N/m)

Consequences and Applications of Surface Tension:

Factors affecting surface tension:


Viscosity

Viscosity is a measure of a liquid's resistance to flow. It describes the internal friction between adjacent layers of a liquid that are moving relative to each other.

Imagine a liquid flowing in layers. A layer moving faster experiences a retarding force from the slower layer below it, and it exerts an accelerating force on the slower layer. This internal resistance is viscosity.

Illustration of viscous flow in layers

Coefficient of Viscosity ($\eta$ or $\mu$): It is a quantitative measure of viscosity. It is defined as the tangential force per unit area required to maintain a unit velocity gradient between two parallel layers of the liquid.

Consider two parallel layers of a liquid separated by a distance $dz$, moving with velocities $v$ and $v+dv$. The velocity gradient is $dv/dz$. The viscous force (F) between these layers is given by Newton's law of viscosity:

$ F = \eta A \frac{dv}{dz} $

Where A is the area of the layers and $\eta$ is the coefficient of viscosity.

The SI unit of viscosity is Pascal-second (Pa s) or N s/m$^2$. The CGS unit is poise (P) (1 poise = 0.1 Pa s). Often, centipoise (cP) is used (1 cP = 1 mPa s).

Factors affecting viscosity:

  1. Intermolecular Forces: Stronger intermolecular forces lead to higher viscosity. Liquids like honey or glycerine have higher viscosity than water or alcohol due to stronger intermolecular forces (like hydrogen bonding) and possibly larger molecular size/complexity causing more friction.
  2. Temperature: Viscosity of liquids decreases significantly with increasing temperature. As temperature increases, the kinetic energy of molecules increases, allowing them to overcome the intermolecular forces more easily and slide past each other, reducing internal friction.
  3. Pressure: The viscosity of liquids generally increases with increasing pressure, as particles are forced closer together, increasing intermolecular interactions.
  4. Molecular Size and Shape: Larger and more complex molecules tend to have higher viscosity.

Viscosity is important in understanding the flow of liquids, lubrication, and transport phenomena in fluids.