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Thermal Expansion and Ideal Gases



Ideal-Gas Equation And Absolute Temperature ($ PV = nRT $)

The behavior of gases can be described by relationships between their macroscopic properties: pressure ($P$), volume ($V$), and temperature ($T$). For gases at low densities and pressures, these properties are related by a simple equation known as the ideal-gas equation. This equation is based on the concept of an ideal gas, which is a theoretical model that approximates the behavior of real gases under certain conditions.


Ideal Gas Model

An ideal gas is defined based on the following assumptions:

Real gases behave approximately as ideal gases at low pressures and high temperatures, where the molecules are far apart and their interactions are minimal.


Empirical Gas Laws

Before the ideal-gas equation was formulated, scientists discovered several empirical laws relating the properties of gases at constant conditions:


The Ideal-Gas Equation

Combining these empirical laws leads to the ideal-gas equation, which relates the pressure, volume, temperature, and the amount of gas (in moles):

$ PV = nRT $

where:

The amount of gas $n$ can also be expressed as $n = N/N_A$, where $N$ is the total number of molecules and $N_A$ is Avogadro's number ($N_A \approx 6.022 \times 10^{23} \, \text{mol}^{-1}$). Substituting this into the ideal-gas equation gives:

$ PV = \frac{N}{N_A} RT $

$ PV = N \left(\frac{R}{N_A}\right) T $

The ratio $R/N_A$ is another fundamental constant called the Boltzmann constant ($k_B$ or $k$). $ k_B = \frac{R}{N_A} \approx \frac{8.314 \, \text{J/(mol} \cdot \text{K})}{6.022 \times 10^{23} \, \text{mol}^{-1}} \approx 1.381 \times 10^{-23} \, \text{J/K} $. The ideal-gas equation can also be written in terms of the number of molecules $N$ and the Boltzmann constant:

$ PV = N k_B T $


Absolute Temperature Scale (Kelvin)

The ideal-gas equation (and the empirical gas laws it is based on) requires the use of an absolute temperature scale, such as the Kelvin scale. If we use Celsius or Fahrenheit scales, the simple proportionalities (like Charles's Law, $V \propto T$) would not hold because these scales have arbitrary zero points.

Absolute zero (0 K) is the theoretical minimum temperature where an ideal gas would exert zero pressure (at constant volume) or occupy zero volume (at constant pressure). This corresponds to the state where the average kinetic energy of the gas molecules is minimum. On the Celsius scale, absolute zero is approximately -273.15°C.

The Kelvin scale is directly related to the average kinetic energy of the gas particles. In the kinetic theory of gases, the average translational kinetic energy per molecule of an ideal gas is given by $\frac{3}{2} k_B T$. This direct proportionality to the absolute temperature $T$ highlights the fundamental nature of the Kelvin scale.

Example 1. Calculate the volume occupied by 2 moles of an ideal gas at a pressure of $1.01 \times 10^5$ Pa and a temperature of 27°C. (Take $R = 8.314 \, \text{J/(mol} \cdot \text{K})$).

Answer:

Amount of gas, $n = 2$ mol.

Absolute pressure, $P = 1.01 \times 10^5$ Pa.

Temperature, $T_C = 27^\circ\text{C}$. Convert to Kelvin: $T_K = T_C + 273.15 = 27 + 273.15 = 300.15$ K. (Use 300 K for simplicity as per common practice in such problems if not specified otherwise).

Ideal gas constant, $R = 8.314 \, \text{J/(mol} \cdot \text{K})$.

Using the ideal-gas equation, $PV = nRT$. Rearrange to solve for $V$:

$ V = \frac{nRT}{P} $

Using $T = 300$ K:

$ V = \frac{(2 \text{ mol}) \times (8.314 \, \text{J/(mol} \cdot \text{K)}) \times (300 \text{ K})}{1.01 \times 10^5 \text{ Pa}} $

$ V = \frac{2 \times 8.314 \times 300}{1.01 \times 10^5} \frac{\text{J}}{\text{Pa}} $ (Units: $\frac{\text{J}}{\text{Pa}} = \frac{\text{N} \cdot \text{m}}{\text{N/m}^2} = \text{N} \cdot \text{m} \cdot \text{m}^2/\text{N} = \text{m}^3$)

$ V = \frac{4988.4}{101000} \text{ m}^3 \approx 0.04939 $ m$^3$.

Using $T = 300.15$ K:

$ V = \frac{2 \times 8.314 \times 300.15}{1.01 \times 10^5} = \frac{4990.8}{101000} \approx 0.04941 $ m$^3$.

The volume occupied by the gas is approximately 0.0494 m$^3$ or 49.4 litres (since 1 m$^3 = 1000$ litres). This is close to the molar volume of an ideal gas at STP (Standard Temperature and Pressure, 0°C and 1 atm), which is 22.4 litres per mole. Here, the temperature is higher than STP, so the volume is larger per mole.



Thermal Expansion

Most substances (solids, liquids, and gases) tend to expand when their temperature increases and contract when their temperature decreases. This phenomenon is called thermal expansion. It is a consequence of the increased average kinetic energy of the particles at higher temperatures, leading to larger average distances between them.


Thermal Expansion in Solids

In solids, particles are held in relatively fixed positions, vibrating about their equilibrium points. As temperature increases, the amplitude of these vibrations increases, leading to a slightly larger average separation between particles and thus expansion of the solid. Solids can expand in length, area, or volume.


Linear Expansion

Linear expansion refers to the change in length of a solid body with temperature. For a solid rod of original length $L_0$ at temperature $T_0$, the change in length $\Delta L$ when the temperature changes by $\Delta T$ is approximately proportional to the original length and the temperature change.

$ \Delta L \propto L_0 \Delta T $

Introducing the constant of proportionality, $\alpha$, called the coefficient of linear expansion:

$ \Delta L = \alpha L_0 \Delta T $

or $ L = L_0 (1 + \alpha \Delta T) $, where $L = L_0 + \Delta L$ is the final length.

The coefficient of linear expansion $\alpha$ is a property of the material. Its units are per degree Celsius (°C$^{-1}$) or per Kelvin (K$^{-1}$). Since $\Delta T$ is the same in °C and K, the value of $\alpha$ is the same in both units. $\alpha$ is generally positive, indicating expansion with increasing temperature, but some materials can have negative $\alpha$ over certain temperature ranges. The value of $\alpha$ can also vary slightly with temperature.


Area Expansion (Superficial Expansion)

Area expansion refers to the change in the surface area of a solid body with temperature. For a solid with original area $A_0$ at temperature $T_0$, the change in area $\Delta A$ for a temperature change $\Delta T$ is approximately:

$ \Delta A = \beta A_0 \Delta T $

where $\beta$ is the coefficient of area expansion. For isotropic solids, $\beta \approx 2\alpha$.

$ A = A_0 (1 + \beta \Delta T) \approx A_0 (1 + 2\alpha \Delta T) $


Volume Expansion (Cubical Expansion)

Volume expansion refers to the change in the volume of a solid, liquid, or gas with temperature. For a substance with original volume $V_0$ at temperature $T_0$, the change in volume $\Delta V$ for a temperature change $\Delta T$ is approximately:

$ \Delta V = \gamma V_0 \Delta T $

where $\gamma$ is the coefficient of volume expansion. For isotropic solids, $\gamma \approx 3\alpha$. For liquids and gases, only volume expansion is typically considered (they don't have fixed shapes). Liquids generally expand more than solids for the same temperature change. For ideal gases, $\gamma$ is related to temperature (see below).

$ V = V_0 (1 + \gamma \Delta T) $

The units of $\beta$ and $\gamma$ are also °C$^{-1}$ or K$^{-1}$.


Thermal Expansion in Liquids

Liquids also expand with increasing temperature. Since liquids do not have a fixed shape, only volume expansion is relevant. Liquids generally have higher coefficients of volume expansion than solids, meaning they expand more significantly for the same temperature change.

When measuring the volume expansion of a liquid in a container, one must consider both the expansion of the liquid and the expansion of the container. The observed (apparent) expansion of the liquid is the difference between the real expansion of the liquid and the expansion of the container.

An important anomaly is the behaviour of water. Water contracts upon heating from 0°C to 4°C (negative thermal expansion) and then expands upon further heating above 4°C. Water has maximum density at 4°C. This anomalous expansion of water is crucial for aquatic life in cold climates, as ice forms on the surface of lakes and rivers (less dense than water) while the denser 4°C water remains at the bottom, allowing life to survive.


Thermal Expansion in Gases

Gases exhibit significant volume expansion with temperature. For an ideal gas at constant pressure, Charles's Law ($V \propto T$) implies thermal expansion. From $PV = nRT$, if P and n are constant, $V = (\frac{nR}{P})T$. This is in the form $V = (\text{constant}) \times T$.

If we use the formula $V = V_0 (1 + \gamma \Delta T)$ and consider an ideal gas at constant pressure $P_0$ and temperature $T_0$ (in Kelvin), then $V_0 = \frac{nRT_0}{P_0}$. At temperature $T = T_0 + \Delta T$, the volume is $V = \frac{nRT}{P_0} = \frac{nR(T_0 + \Delta T)}{P_0} = \frac{nRT_0}{P_0} (1 + \frac{\Delta T}{T_0}) = V_0 (1 + \frac{\Delta T}{T_0})$.

Comparing $V = V_0 (1 + \gamma \Delta T)$ with $V = V_0 (1 + \frac{\Delta T}{T_0})$, we see that for an ideal gas at constant pressure, the coefficient of volume expansion is $\gamma = \frac{1}{T_0}$, where $T_0$ is the initial absolute temperature. This means the thermal expansion coefficient for gases is not a constant like for solids and liquids but depends on the initial temperature. At higher temperatures, the fractional volume change for a given temperature change is smaller.


Applications and Consequences of Thermal Expansion

Example 2. A steel rod has a length of 1.5 m at 20°C. The coefficient of linear expansion of steel is $12 \times 10^{-6} \, ^\circ\text{C}^{-1}$. What is the increase in length of the rod when its temperature is raised to 70°C?

Answer:

Original length, $L_0 = 1.5$ m.

Initial temperature, $T_0 = 20^\circ\text{C}$.

Final temperature, $T = 70^\circ\text{C}$.

Change in temperature, $\Delta T = T - T_0 = 70^\circ\text{C} - 20^\circ\text{C} = 50^\circ\text{C}$.

Coefficient of linear expansion of steel, $\alpha = 12 \times 10^{-6} \, ^\circ\text{C}^{-1}$.

The increase in length $\Delta L$ is given by the formula for linear expansion:

$ \Delta L = \alpha L_0 \Delta T $

$ \Delta L = (12 \times 10^{-6} \, ^\circ\text{C}^{-1}) \times (1.5 \text{ m}) \times (50 \, ^\circ\text{C}) $

$ \Delta L = (12 \times 1.5 \times 50) \times 10^{-6} $ m

$ \Delta L = (18 \times 50) \times 10^{-6} $ m

$ \Delta L = 900 \times 10^{-6} $ m

$ \Delta L = 9 \times 10^2 \times 10^{-6} $ m $= 9 \times 10^{-4}$ m.

The increase in length of the steel rod is $9 \times 10^{-4}$ metres, which is 0.9 mm.