Thermodynamic state variables and equation of state

This topic sits inside the CBSE Class 11 Physics Thermodynamics chapter, right after thermal equilibrium, the zeroth law and the first law ($\Delta Q = \Delta U + \Delta W$), and just before the chapter turns to naming specific processes — isothermal, adiabatic, isochoric, isobaric and cyclic. The ideal gas equation of state $PV = nRT$ itself is developed in full, from kinetic theory, in the separate Kinetic Theory chapter, but the Thermodynamics chapter uses it constantly as a tool: every $P$–$V$ diagram, every "find the work done as the gas expands" question, leans on this single relation. You are expected to identify $P$, $V$, $T$, $n$ as the basic state variables, tell extensive from intensive variables, and apply $PV = nRT$ and the combined gas law fluently before tackling the process-specific work and heat calculations that follow.

A gas sample settles into equilibrium when its pressure, volume and temperature stop changing on their own — stir it, heat it unevenly, and wait, and eventually every part of the gas agrees on a single $P$, a single $T$. Once it has settled, the gas is completely described by a short list of numbers:

• Pressure $P$ — the force per unit area the gas exerts on its container walls, SI unit pascal (Pa), $1\ \text{Pa} = 1\ \text{N/m}^2$.
• Volume $V$ — the space the gas occupies, SI unit cubic metre (m³); the litre ($1\ \text{L} = 10^{-3}\ \text{m}^3$) is a common lab unit.
• Temperature $T$ — measured in kelvin (K) for all gas-law work; it tracks the average kinetic energy of the molecules (developed fully in Kinetic Theory).
• Amount of substance $n$ — the number of moles of gas present, or equivalently the mass $m$ divided by the molar mass $M$.

These four are called state variables or state functions precisely because each one has a definite value once the gas has settled into equilibrium — it does not matter whether the gas arrived there by slow heating, sudden compression, or any other route. Contrast this with quantities like heat $Q$ and work $W$, which are not state variables: they depend on the path taken between two states, not just on the states themselves. State variables further split by how they respond to scaling up the system. Take two identical boxes of the same gas at the same $P$ and $T$, and merge them into one box (doubling the amount of gas at the same $P$, $T$):

• Extensive variables double when you merge the two boxes — volume $V$, internal energy $U$, and total mass all double. They depend on how much gas you have.
• Intensive variables stay the same when you merge the boxes — pressure $P$, temperature $T$, and density $\rho$ are unchanged. They describe the condition of the gas, not its quantity.

An equation of state is the mathematical rule connecting a system's state variables in equilibrium. For an ideal gas — the model CBSE Class 11 uses throughout this unit — that rule is

where: $P$ is pressure (Pa), $V$ is volume (m³), $n$ is the number of moles, $T$ is absolute temperature (K), and $R = 8.31\ \text{J mol}^{-1}\text{K}^{-1}$ is the universal gas constant, the same number for every ideal gas regardless of what it is made of. Written in terms of the total mass $m$ and molar mass $M$ of the gas (using $n = m/M$), the same law reads

where: this second form is convenient when you are told a mass of gas rather than a mole count. The physical reading of $PV = nRT$ is that once you fix $n$ (the amount of gas you have sealed in), only two of $P$, $V$, $T$ are yours to set independently — the third is then forced. Push the piston in (lower $V$) at constant $T$ and $P$ must rise; heat the gas (raise $T$) in a rigid box and $P$ must rise; and so on. This is exactly what makes the state of a fixed amount of gas representable as a single point on a $P$–$V$ diagram, with the temperature implied by that point through the equation of state.

Because $P$, $V$, $T$ are state variables, only the endpoints of a path on this diagram matter for computing the change in internal energy $\Delta U$ — but the shape of the path itself determines how much heat $Q$ flows and how much work $W$ is done, a distinction the first law of thermodynamics will make precise. Real gases follow $PV = nRT$ only approximately, deviating more at high pressure and low temperature where molecules start to interact and take up appreciable volume — but CBSE Class 11 treats every gas in this unit as ideal.

Board level - volume of a gas from its state variables

A cylinder contains 2 moles of an ideal gas at pressure $2\times10^5$ Pa and temperature 300 K. Find the volume occupied by the gas. ($R = 8.31\ \text{J mol}^{-1}\text{K}^{-1}$.)

Step 1. Identify the state variables given: $n = 2\ \text{mol}$, $P = 2\times10^5\ \text{Pa}$, $T = 300\ \text{K}$. The unknown is $V$, and since all three other quantities in $PV = nRT$ are known, the equation of state alone is enough — no combined gas law needed.

Step 2. Rearrange $PV = nRT$ for volume: $V = \dfrac{nRT}{P}$.

Step 3. Substitute the numbers: $V = \dfrac{2 \times 8.31 \times 300}{2\times10^5} = \dfrac{4986}{2\times10^5}$.

Step 4. Compute and convert to a familiar unit: $V = 0.02493\ \text{m}^3 \approx 2.49\times10^{-2}\ \text{m}^3$. Since $1\ \text{m}^3 = 1000\ \text{L}$, this is about 24.9 litres — roughly the volume of a large cooking-gas cylinder, which is a useful sanity check on the order of magnitude.

Competitive level - verifying the equation of state across two states

A fixed mass of gas is initially at $P_1 = 1.2\times10^5$ Pa, $V_1 = 3\times10^{-3}\ \text{m}^3$, $T_1 = 290$ K. It is heated and allowed to expand until its final state is $P_2 = 1.5\times10^5$ Pa, $V_2 = 4\times10^{-3}\ \text{m}^3$. Find the final temperature $T_2$, and verify that the same equation of state is satisfied at both states with the same $n$.

Step 1. Since the number of moles is fixed (no gas added or removed), apply the combined gas law derived above: $\dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2}$, so $T_2 = T_1 \times \dfrac{P_2V_2}{P_1V_1}$.

Step 2. Substitute the numbers: $T_2 = 290 \times \dfrac{(1.5\times10^5)(4\times10^{-3})}{(1.2\times10^5)(3\times10^{-3})} = 290 \times \dfrac{600}{360}$.

Step 3. Compute: $T_2 = 290 \times 1.667 = 483.3\ \text{K}$. This is the temperature the gas must reach for its pressure and volume to end up at the stated final values.

Step 4. Cross-check by finding $n$ independently from each state. From state 1: $n = \dfrac{P_1V_1}{RT_1} = \dfrac{(1.2\times10^5)(3\times10^{-3})}{8.31\times290} = \dfrac{360}{2409.9} = 0.1494\ \text{mol}$. From state 2, using the just-found $T_2$: $n = \dfrac{P_2V_2}{RT_2} = \dfrac{(1.5\times10^5)(4\times10^{-3})}{8.31\times483.3} = \dfrac{600}{4016.2} = 0.1494\ \text{mol}$. The two values of $n$ agree exactly, so $T_2 = 483.3\ \text{K}$, and both states obey $PV = nRT$ with the same fixed $n \approx 0.1494\ \text{mol}$, as they must. Timing note: In an exam, do not compute $n$ twice as done here for verification — that is a check, not a requirement. For a straight numerical answer, apply the combined gas law directly (Steps 1–3) and stop; that should take under two minutes, leaving the verification step only for when the question explicitly asks you to confirm consistency or find $n$.

• Forgetting to convert temperature to kelvin. The equation of state requires absolute temperature throughout; plugging in a Celsius value directly (e.g. using $27$ instead of $300\ \text{K}$) is one of the most common marks-losing errors in this topic.
• Treating $P$, $V$, $T$ as three independent dials. For a fixed amount of gas, only two of the three can be chosen freely — the third is always forced by $PV = nRT$. A question that seems to give you three independent conditions on the same fixed sample is either testing this trap or giving redundant (consistent) information.
• Mixing up extensive and intensive variables. Pressure and temperature do not change just because you have "more system" at fixed conditions elsewhere — but they absolutely do change if you add gas at fixed $V$ and $T$ (doubling $n$ at fixed $V, T$ doubles $P$, by $PV = nRT$). The trap is applying "intensive means unaffected by anything" too broadly; it only means unaffected by scaling the system size at the same conditions.
• Unit mismatches. $R = 8.31\ \text{J mol}^{-1}\text{K}^{-1}$ demands pressure in pascals and volume in m³; using litres or atmospheres without converting first is a very common arithmetic slip, especially under time pressure.
• Confusing a state variable with a path-dependent quantity. $\Delta U$ depends only on the initial and final states (since $U$ is a state variable), but $Q$ and $W$ depend on the specific path taken between those states — do not assume that because $P_1V_1T_1$ and $P_2V_2T_2$ match between two problems, the heat and work involved must also match.

• List the basic thermodynamic state variables ($P$, $V$, $T$, $n$) and explain why they are called state variables.
• Classify a given quantity as extensive or intensive by imagining what happens when you double the amount of the system.
• State and apply the ideal gas equation of state $PV = nRT$, in both mole and mass forms.
• Derive and apply the combined gas law $\dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2}$ for a fixed amount of gas moving between two states.
• Explain why $P$, $V$, $T$ cannot all vary independently for a fixed amount of ideal gas.
• Represent a gas's equilibrium state as a point on a $P$–$V$ diagram, and a process as a path between two such points, distinguishing state variables (path-independent) from heat and work (path-dependent).

Ask the tutor: Want more practice converting between litres/atm and SI units in $PV = nRT$, or a walkthrough of how this equation of state sets up the isothermal and adiabatic processes coming up next? Ask the AI tutor for extra worked problems.