Gas laws graphical interpretation — PV, PT, VT diagrams reading

Question

Identify the gas law from each graph type: (a) PV vs P at constant T, (b) V vs T at constant P, (c) P vs T at constant V. Sketch each graph and explain what the slope or intercept tells us.

(CBSE 11 + JEE Main pattern)

Solution — Step by Step

From Boyle’s law: $PV = k$ (constant) at fixed T.

A plot of $PV$ vs $P$ is a horizontal straight line — the product $PV$ stays constant regardless of pressure.

A plot of $P$ vs $V$ gives a rectangular hyperbola ( $P = k/V$ ).

A plot of $P$ vs $1/V$ gives a straight line through the origin with slope $= k = nRT$ .

From Charles’s law: $V = \frac{nR}{P} T$ at fixed P.

A plot of $V$ vs $T$ (in Kelvin) is a straight line through the origin with slope $= nR/P$ .

At higher pressure, the slope is smaller (same amount of gas occupies less volume per degree of temperature).

If plotted against Celsius, the line intercepts the x-axis at $-273.15°$ C (absolute zero) — this is how absolute zero was first estimated.

$P = \frac{nR}{V} T$ at fixed V.

A plot of $P$ vs $T$ (Kelvin) is a straight line through the origin with slope $= nR/V$ .

At larger volume, the slope is smaller. The line also extrapolates to zero pressure at 0 K.

flowchart TD
    A["Gas Law Graphs"] --> B["PV vs P (const T)"]
    A --> C["V vs T (const P)"]
    A --> D["P vs T (const V)"]
    B --> E["Horizontal line: PV = constant"]
    C --> F["Straight line through origin"]
    C --> G["Slope = nR/P"]
    D --> H["Straight line through origin"]
    D --> I["Slope = nR/V"]
    F --> J["Higher P → smaller slope"]
    I --> K["Higher V → smaller slope"]

Why This Works

The ideal gas equation $PV = nRT$ contains three variables (P, V, T). When we fix one, the relationship between the other two becomes simple. Boyle’s law fixes T, Charles’s law fixes P, and Gay-Lussac’s law fixes V. Each gives a linear relationship (or inverse relationship) that produces characteristic graph shapes.

The straight-line graphs passing through the origin tell us that these are direct proportionalities. The slope contains information about the amount of gas ( $n$ ) and the fixed variable. By comparing slopes of different lines on the same graph, we can determine which line corresponds to higher or lower values of the fixed variable.

Alternative Method — Using the Ideal Gas Equation Directly

For any unknown gas law graph, start with $PV = nRT$ and rearrange:

Want $V$ vs $T$ ? → $V = (nR/P)T$ → slope is $nR/P$
Want $P$ vs $1/V$ ? → $P = nRT \cdot (1/V)$ → slope is $nRT$
Want $PV$ vs $T$ ? → $PV = nRT$ → slope is $nR$ (identity of the gas determines this)

For JEE MCQs, graph-based questions often show multiple lines and ask which line represents a higher temperature or pressure. Remember: for V vs T at constant P, the line with the SMALLER slope represents HIGHER pressure (because slope $= nR/P$ , and larger P means smaller slope). Students who memorise without understanding get confused by these comparison questions.

Common Mistake

Students plot $V$ vs $T$ in Celsius and expect a straight line through the origin. The line passes through the origin ONLY when temperature is in Kelvin. In Celsius, it passes through $(−273.15, 0)$ , not $(0, 0)$ . Using Celsius in the ideal gas equation gives wrong answers. Always convert to Kelvin first: $T(K) = T(°C) + 273.15$ .