Gas Graphs — Concepts, Formulas & Examples

Gas laws are easier to remember as graphs than as equations. CBSE Class 11 and NEET test graph-reading questions frequently — identify the law from the shape, predict the trend, calculate from the plot. If you can sketch the six key gas law graphs from memory, you can handle any question on this topic.

Why graphs and not just equations? Because examiners love giving you a curve and asking which law it represents, or giving two curves and asking which gas has higher temperature. The shape of the graph encodes the physics — a straight line means direct proportionality, a hyperbola means inverse proportionality, and the intercept tells you a physical constant.

Core Concepts

Boyle’s law (P-V relationship at constant T)

At constant temperature and amount of gas, $PV = \text{constant}$ , or $P \propto 1/V$ .

Graph shapes:

P vs V — rectangular hyperbola (as V increases, P decreases, but the product PV stays constant)
P vs 1/V — straight line through the origin (direct proportionality)
PV vs P — horizontal line (PV is constant, independent of P)
log P vs log V — straight line with slope $-1$

Each curve at a different temperature is called an isotherm. Higher temperature isotherms sit further from the origin on the P vs V graph because at higher T, the gas occupies more volume at any given pressure.

P_1V_1 = P_2V_2 \quad \text{(at constant T, n)}

This is the workhorse equation. Given any three values, you find the fourth.

Charles’s law (V-T relationship at constant P)

At constant pressure and amount of gas, $V \propto T$ (T must be in Kelvin).

Graph shapes:

V vs T (Kelvin) — straight line through the origin
V vs T (Celsius) — straight line intercepting the temperature axis at $-273.15°\text{C}$ (absolute zero)

The intercept on the temperature axis is the experimental basis for absolute zero. All gases, when extrapolated, converge to zero volume at $-273.15°\text{C}$ — though in practice they liquefy first.

Lines at different pressures are called isobars. A higher pressure isobar has a smaller slope because the same temperature increase produces a smaller volume change when the gas is more compressed.

\frac{V_1}{T_1} = \frac{V_2}{T_2} \quad \text{(at constant P, n)}

Always use Kelvin. This is the single most common calculation mistake in gas law problems.

Gay-Lussac’s law (P-T relationship at constant V)

At constant volume and amount of gas, $P \propto T$ .

Graph shapes:

P vs T (Kelvin) — straight line through the origin
P vs T (Celsius) — straight line intercepting the temperature axis at $-273.15°\text{C}$

Lines at different volumes are called isochores (or isometrics). A smaller volume isochore has a steeper slope — the same temperature increase produces a larger pressure change in a smaller container.

\frac{P_1}{T_1} = \frac{P_2}{T_2} \quad \text{(at constant V, n)}

Combined gas law

When temperature, pressure and volume all change simultaneously:

\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}

This combines Boyle, Charles and Gay-Lussac. Use it when more than one variable changes between two states.

Ideal gas equation

PV = nRT

where $R$ is the universal gas constant:

$R = 8.314$ J mol $^{-1}$ K $^{-1}$ (SI)
$R = 0.0821$ L atm mol $^{-1}$ K $^{-1}$ (commonly used in gas law problems)
$R = 2$ cal mol $^{-1}$ K $^{-1}$ (approximate, useful in thermochemistry)

At STP (273.15 K, 1 atm), one mole of any ideal gas occupies 22.4 L — this is the molar volume.

PV = nRT

This is the most powerful single equation in gas chemistry. Every other gas law is a special case of this with one or more variables held constant.

Real gas deviations and the compressibility factor

Ideal gas assumes no molecular volume and no intermolecular forces. Real gases deviate at high pressure and low temperature.

The compressibility factor $Z = \frac{PV}{nRT}$ measures deviation:

$Z = 1$ — ideal behaviour
$Z < 1$ — attractive forces dominate (gas is easier to compress than ideal), common at moderate pressures
$Z > 1$ — molecular volume dominates (gas is harder to compress than ideal), common at very high pressures

Graph of Z vs P:

For an ideal gas, $Z = 1$ at all pressures (horizontal line)
For real gases, the curve dips below 1 at moderate P then rises above 1 at high P
$\text{H}_2$ and He show $Z > 1$ at almost all pressures because their attractive forces are negligible

Van der Waals equation

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

The $a$ term corrects for intermolecular attractions. The $b$ term corrects for molecular volume. Gases with larger, more polar molecules have larger $a$ values. Gases with larger molecules have larger $b$ values.

\left(P + \frac{a}{V_m^2}\right)(V_m - b) = RT

where $V_m = V/n$ is the molar volume. The constants $a$ and $b$ are specific to each gas.

Dalton’s law of partial pressures

In a mixture of non-reacting gases, the total pressure equals the sum of the partial pressures:

P_{\text{total}} = P_1 + P_2 + P_3 + \cdots

Each partial pressure is $P_i = x_i \cdot P_{\text{total}}$ , where $x_i$ is the mole fraction of gas $i$ .

Worked Examples

If you see a P vs V graph that is a rectangular hyperbola, you are looking at Boyle’s law at constant temperature. To find any unknown state from a known state, use $P_1V_1 = P_2V_2$ . For example, if a gas at 2 atm occupies 5 L, then at 5 atm it occupies $(2 \times 5)/5 = 2$ L.

On a P vs V graph, two isotherms are shown — one closer to the origin and one further away. The isotherm further from the origin corresponds to the higher temperature. Why? At higher T, any given pressure corresponds to a larger volume, pushing the curve outward.

A gas occupies 500 mL at 27°C and 2 atm. Find its volume at STP.

Using the combined gas law: $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$

$\frac{2 \times 500}{300} = \frac{1 \times V_2}{273}$

$V_2 = \frac{2 \times 500 \times 273}{300 \times 1} = \frac{273000}{300} = 910$ mL

Note: we converted 27°C to 300 K and used STP values of 1 atm and 273 K.

A container has 0.5 mol $\text{N}_2$ and 0.3 mol $\text{O}_2$ at a total pressure of 4 atm. Find the partial pressure of $\text{O}_2$ .

Total moles = 0.5 + 0.3 = 0.8. Mole fraction of $\text{O}_2$ = 0.3/0.8 = 0.375.

$P_{\text{O}_2} = 0.375 \times 4 = 1.5$ atm.

1 mol of a gas at 300 K and 20 atm occupies 1.2 L. Is it behaving ideally?

$Z = \frac{PV}{nRT} = \frac{20 \times 1.2}{1 \times 0.0821 \times 300} = \frac{24}{24.63} = 0.974$

$Z < 1$ , so attractive forces dominate. The gas is slightly easier to compress than an ideal gas. This is typical behaviour at moderate pressures.

Common Mistakes

Using Celsius in gas law calculations. Always convert to Kelvin by adding 273. Using 25°C instead of 298 K will give a completely wrong answer and is the single most frequent error in gas law numericals.

Confusing P vs V (hyperbola) with P vs 1/V (straight line). Both represent Boyle’s law, but the graph shapes are different. If the question asks to identify the law from a straight line through the origin, check the axes carefully — it could be Boyle’s (P vs 1/V), Charles’s (V vs T) or Gay-Lussac’s (P vs T).

Assuming real gases follow the ideal gas law exactly. They deviate at extreme conditions. At high pressures, molecular volume matters ( $Z > 1$ ). At moderate pressures, attractions matter ( $Z < 1$ ). Only at low pressure and high temperature does ideal behaviour hold well.

Forgetting that 22.4 L is the molar volume at STP only. At room temperature (25°C, 1 atm), the molar volume is about 24.5 L. Using 22.4 L at non-STP conditions is a common numerical error.

Mixing up the slope interpretations on V vs T graphs. A steeper line on a V vs T(K) plot at constant P does NOT mean higher pressure — it means lower pressure (because at lower P, the gas is more expandable, so V changes more per kelvin).

Exam Weightage and Strategy

Gas laws are part of the States of Matter chapter in CBSE Class 11, carrying 4-5 marks in boards. NEET asks 1-2 questions, usually graph-based or calculation-based. JEE Main goes deeper into real gas behaviour and the Van der Waals equation. For all three exams, graph identification is the most common question format.

The PYQ pattern for gas graphs is remarkably consistent:

CBSE boards: Sketch the graph of P vs V at two different temperatures. Which isotherm is at higher T? (2-3 marks)
NEET: Given a Z vs P graph, identify which gas deviates the most. Or a simple numerical using the ideal gas equation.
JEE Main: Calculate the compressibility factor, or use Van der Waals to explain why a real gas liquefies below a certain temperature.

Sketch six graphs on one page — P vs V, P vs 1/V, PV vs P (Boyle’s law), V vs T(K), V vs T(°C) (Charles’s law), and P vs T(K) (Gay-Lussac’s law). That single page handles all PYQs on gas law graphs. Add a Z vs P curve for JEE preparation.

Practice Questions

Q1. A gas at 27°C and 1 atm occupies 10 L. What volume will it occupy at 127°C and 2 atm?

Using the combined gas law: $V_2 = V_1 \times \frac{P_1}{P_2} \times \frac{T_2}{T_1} = 10 \times \frac{1}{2} \times \frac{400}{300} = 6.67$ L. (27°C = 300 K, 127°C = 400 K)

Q2. On a P vs V graph, how can you tell which isotherm is at higher temperature?

The isotherm at higher temperature is farther from the origin. At any given pressure, the higher-temperature gas has a larger volume, so the curve is shifted outward (to the right and upward).

Q3. Why does $\text{H}_2$ show $Z > 1$ at almost all pressures?

$\text{H}_2$ has very weak intermolecular attractions (small $a$ value) due to its tiny, non-polar molecules. The molecular volume effect ( $b$ term) dominates even at moderate pressures, making the gas harder to compress than ideal. Therefore $Z = PV/nRT > 1$ at nearly all pressures.

Q4. What happens to the V vs T(°C) graph if we use a different gas at the same pressure?

The straight line still intercepts the temperature axis at $-273.15°$ C (absolute zero). The slope may differ if the amount of gas differs, but the x-intercept is always the same for all ideal gases. This is the experimental basis for the existence of absolute zero.

Q5. Calculate the density of $\text{CO}_2$ at STP.

At STP, 1 mol of any gas occupies 22.4 L. Molar mass of $\text{CO}_2$ = 44 g/mol. Density $= \frac{M}{V_m} = \frac{44}{22.4} = 1.96$ g/L. This is why $\text{CO}_2$ sinks to the ground — it is denser than air (average molar mass ~29 g/mol, density ~1.29 g/L at STP).

FAQs

Why do we need to use Kelvin and not Celsius in gas law calculations?

Gas laws describe proportional relationships (V proportional to T, P proportional to T). These proportionalities only hold when T is measured from absolute zero. In Celsius, 0°C is not the absence of thermal energy — it is an arbitrary reference. Using Celsius would give $V \propto (T + 273)$ , which is not a clean proportion. Kelvin avoids this by starting at true zero.

What is the difference between ideal and real gases?

An ideal gas has molecules with zero volume and zero intermolecular forces. Real gas molecules have finite size and attract each other. At low pressure and high temperature, all gases behave nearly ideally because molecules are far apart. At high pressure or low temperature, deviations become significant and we need the Van der Waals equation.

Why does PV vs P give a horizontal line for an ideal gas?

From $PV = nRT$ , if T and n are constant, then $PV = \text{constant}$ . Plotting this constant against P gives a horizontal line. For real gases, the PV vs P curve dips and then rises, reflecting the competing effects of attraction (which decreases PV) and molecular volume (which increases PV).

What is Boyle temperature?

The temperature at which a real gas obeys Boyle’s law over a significant range of pressures. At this temperature, the attractive and repulsive effects cancel, and the gas behaves ideally. Boyle temperature $T_B = a/(Rb)$ from Van der Waals constants. For $\text{N}_2$ , $T_B \approx 327°\text{C}$ .

Gas law graphs are pattern recognition. Once you know the shapes, the law is obvious at a glance, and the numerical is just substitution.