Equation of state of a perfect gas

In physics, an equation of state is a thermodynamic equation relating the pressure, temperature, and volume of a system in thermodynamic equilibrium. For a perfect gas, this relationship is described by a fundamental equation that combines the gas laws. A perfect gas, also known as an ideal gas, is a theoretical gas composed of many non-interacting point particles, each of which has zero rest mass and is subject to no forces other than collisions.

Boyle's Law

Robert Boyle discovered in 1662 that for a fixed amount of gas at a constant temperature, the pressure and volume are inversely proportional. This means that if you increase the pressure on a gas, its volume decreases proportionally, and vice versa. Mathematically, this can be expressed as:

$P \propto \frac{1}{V}$

or

$PV = constant$

If a gas at pressure $P_1$ and volume $V_1$ changes its pressure to $P_2$ and volume to $V_2$ while the temperature and the amount of gas remain constant, then:

$P_1V_1 = P_2V_2$

Example: Imagine a sealed container of air at room temperature. If you squeeze the container, reducing its volume, the pressure inside the container will increase.

Charles's Law

Jacques Charles discovered in 1787 that for a fixed amount of gas at constant pressure, the volume is directly proportional to its absolute temperature. This means that if you heat a gas, its volume will expand, and if you cool it, its volume will contract. The temperature here must be in Kelvin (absolute temperature).

$V \propto T$

or

$\frac{V}{T} = constant$

If a gas at volume $V_1$ and temperature $T_1$ changes its volume to $V_2$ and temperature to $T_2$ while the pressure and the amount of gas remain constant, then:

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

Example: A balloon filled with air will expand when placed in the sun (temperature increases) and shrink when taken into a cold room (temperature decreases).

Gay-Lussac's Law

Joseph Louis Gay-Lussac discovered in 1808 that for a fixed amount of gas at constant volume, the pressure is directly proportional to its absolute temperature. This means that if you heat a gas in a rigid container, its pressure will increase.

$P \propto T$

or

$\frac{P}{T} = constant$

If a gas at pressure $P_1$ and temperature $T_1$ changes its pressure to $P_2$ and temperature to $T_2$ while the volume and the amount of gas remain constant, then:

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

Example: The pressure inside a car tire increases on a hot day because the air inside heats up, and since the tire's volume is relatively constant, the pressure rises.

Avogadro's Law

Amedeo Avogadro proposed in 1811 that equal volumes of all gases, at the same temperature and pressure, have the same number of molecules. This implies that the volume of a gas is directly proportional to the number of moles (or molecules) of the gas, provided the temperature and pressure are kept constant.

$V \propto n$

or

$\frac{V}{n} = constant$

where $n$ is the number of moles of the gas.

Example: One liter of hydrogen gas at STP (Standard Temperature and Pressure) contains the same number of molecules as one liter of oxygen gas at STP.

The Ideal Gas Law (Perfect Gas Equation)

By combining Boyle's Law, Charles's Law, Gay-Lussac's Law, and Avogadro's Law, we arrive at the Ideal Gas Law. This law describes the state of a hypothetical ideal gas. It states that the product of the pressure and volume of a gas is directly proportional to the product of the number of moles and the absolute temperature.

$PV \propto nT$

Introducing a constant of proportionality, the universal gas constant $R$, we get the equation of state for a perfect gas:

$PV = nRT$

Where:

• $P$ is the absolute pressure of the gas.
• $V$ is the volume of the gas.
• $n$ is the amount of substance of the gas (in moles).
• $R$ is the ideal gas constant, approximately 8.314 J/(mol·K) or 0.0821 L·atm/(mol·K).
• $T$ is the absolute temperature of the gas (in Kelvin).

The value of $R$ is the same for all ideal gases.

The van der Waals Equation (for real gases)

The ideal gas law is a simplification. Real gases deviate from this behavior, especially at high pressures and low temperatures. Johannes van der Waals proposed an equation for real gases that accounts for the finite volume of gas molecules and the attractive forces between them.

The van der Waals equation is:

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

Where:

• $a$ is a constant that corrects for the intermolecular attractive forces. It is larger for gases with stronger attractions.
• $b$ is a constant that corrects for the finite volume occupied by the gas molecules. It is larger for molecules with larger volumes.

The term $\frac{an^2}{V^2}$ represents the reduction in pressure due to intermolecular attractions, and the term $(V - nb)$ represents the reduction in volume available to the gas molecules due to their own volume.

For most competitive exam purposes, the ideal gas law ($PV=nRT$) is sufficient unless specifically asked about real gases or deviations.

Work done on compressing a gas

Work done on or by a gas is a fundamental concept in thermodynamics. When a gas is compressed, work is done *on* the gas, which increases its internal energy (and potentially its temperature or pressure). Conversely, when a gas expands, it does work *on* its surroundings, and this work is done *by* the gas, usually decreasing its internal energy.

Work done in Isobaric Process (Constant Pressure)

An isobaric process is one that occurs at constant pressure. If a gas is compressed from an initial volume $V_1$ to a final volume $V_2$ at a constant pressure $P$, the work done *by* the gas is given by $W = P \Delta V$.

$W_{by \, gas} = P (V_2 - V_1)$

Since $V_2 < V_1$ during compression, $(V_2 - V_1)$ is negative. Therefore, the work done *by* the gas is negative.

The work done *on* the gas ($W_{on \, gas}$) is the negative of the work done *by* the gas.

$W_{on \, gas} = -W_{by \, gas} = -P (V_2 - V_1) = P (V_1 - V_2)$

Since $V_1 > V_2$, $(V_1 - V_2)$ is positive, meaning the work done *on* the gas is positive.

Example: Imagine a cylinder with a piston containing a gas. If you push the piston down slowly, keeping the pressure constant, you are doing work on the gas to compress it.

Work done in Isothermal Process (Constant Temperature)

An isothermal process occurs at a constant temperature. For an ideal gas, if the temperature is constant, then $PV = constant$ (from Boyle's Law). To calculate the work done during compression or expansion in an isothermal process, we need to integrate the pressure over the change in volume.

The work done *by* the gas during an isothermal process from volume $V_1$ to $V_2$ is given by:

$W_{by \, gas} = \int_{V_1}^{V_2} P \, dV$

Since $PV = nRT$, we have $P = \frac{nRT}{V}$. Substituting this into the integral:

$W_{by \, gas} = \int_{V_1}^{V_2} \frac{nRT}{V} \, dV$

Since $n$, $R$, and $T$ are constant in an isothermal process:

$W_{by \, gas} = nRT \int_{V_1}^{V_2} \frac{1}{V} \, dV$

The integral of $\frac{1}{V}$ is $\ln|V|$. So,

$W_{by \, gas} = nRT [\ln|V|]_{V_1}^{V_2} = nRT (\ln V_2 - \ln V_1)$

Using the property of logarithms $\ln a - \ln b = \ln(a/b)$:

$W_{by \, gas} = nRT \ln\left(\frac{V_2}{V_1}\right)$

For compression, $V_2 < V_1$, so $\frac{V_2}{V_1} < 1$. The natural logarithm of a number less than 1 is negative. Thus, the work done *by* the gas is negative, as expected.

The work done *on* the gas during isothermal compression is:

$W_{on \, gas} = -W_{by \, gas} = -nRT \ln\left(\frac{V_2}{V_1}\right)$

Since $V_2 < V_1$, $\frac{V_2}{V_1} < 1$, and $\ln\left(\frac{V_2}{V_1}\right)$ is negative. Therefore, $W_{on \, gas}$ is positive.

We can also express this in terms of pressures. Since $P_1V_1 = P_2V_2$ for an isothermal process, $\frac{V_2}{V_1} = \frac{P_1}{P_2}$. So,

$W_{by \, gas} = nRT \ln\left(\frac{P_1}{P_2}\right)$

And

$W_{on \, gas} = nRT \ln\left(\frac{P_2}{P_1}\right)$

Example: Compressing a gas in a cylinder very slowly, allowing heat to escape so the temperature remains constant.

Work done in Adiabatic Process (No Heat Exchange)

An adiabatic process is one where no heat is exchanged between the system and its surroundings ($Q=0$). In an adiabatic compression, work is done on the gas, and this work increases the internal energy of the gas, leading to a rise in temperature. For an ideal gas, the relationship between pressure and volume in an adiabatic process is given by:

$PV^\gamma = constant$

where $\gamma$ (gamma) is the adiabatic index or heat capacity ratio ($\gamma = C_p/C_v$). $\gamma > 1$.

The work done *by* the gas during an adiabatic process from state 1 ($P_1, V_1$) to state 2 ($P_2, V_2$) is:

$W_{by \, gas} = \frac{P_1V_1 - P_2V_2}{\gamma - 1}$

Alternatively, using the ideal gas law ($PV=nRT$), we can write $P_1V_1 = nRT_1$ and $P_2V_2 = nRT_2$.

$W_{by \, gas} = \frac{nRT_1 - nRT_2}{\gamma - 1} = \frac{nR(T_1 - T_2)}{\gamma - 1}$

For adiabatic compression, $V_2 < V_1$ and $T_2 > T_1$. Thus, $(T_1 - T_2)$ is negative, making $W_{by \, gas}$ negative.

The work done *on* the gas during adiabatic compression is:

$W_{on \, gas} = -W_{by \, gas} = \frac{P_2V_2 - P_1V_1}{\gamma - 1}$

or

$W_{on \, gas} = \frac{nR(T_2 - T_1)}{\gamma - 1}$

Since $T_2 > T_1$, $W_{on \, gas}$ is positive, as expected for work done on a system that increases its internal energy.

Example: Rapid compression of a gas, such as in a diesel engine cylinder, where there isn't enough time for significant heat transfer.

Graphical Representation of Work Done

The work done by or on a gas can be represented graphically on a Pressure-Volume (P-V) diagram. The work done is equal to the area under the curve representing the process on the P-V diagram.

• Isobaric Process: A horizontal line on the P-V diagram. The area is a rectangle: $W = P \Delta V$.
• Isothermal Process: A curve following $PV = constant$. The area under this curve is calculated by integration, resulting in $nRT \ln(V_2/V_1)$.
• Adiabatic Process: A steeper curve than the isothermal curve, following $PV^\gamma = constant$. The area under this curve is $\frac{P_1V_1 - P_2V_2}{\gamma - 1}$.

During compression, the volume decreases ($V_2 < V_1$). If moving from right to left on the P-V diagram, the area represents the work done *on* the gas. For a given change in volume, the adiabatic curve lies above the isothermal curve, indicating higher pressure and thus more work done during adiabatic compression compared to isothermal compression.

Work done in other processes

While isobaric, isothermal, and adiabatic processes are common, work can be calculated for any process where the pressure-volume relationship is known. For example, in a general process where pressure changes as a function of volume, the work done is always given by the integral:

$W = \int_{V_1}^{V_2} P(V) \, dV$

For an isochoric process (constant volume, $\Delta V = 0$), the work done is always zero, as $W = P \Delta V = P \times 0 = 0$. No work is done if the volume of the gas does not change.