Recall and use the equation \$pV = \text{constant}\$ for a fixed mass of gas at constant temperature, and draw a simple graph to show this relationship.
For a given amount of gas kept at a constant temperature, the product of its pressure (\$p\$) and volume (\$V\$) never changes:
\$pV = \text{constant}\$
Think of it like a rubber balloon: if you squeeze it (reduce \$V\$), the air inside gets more crowded, so the pressure rises. If you let it expand, the pressure drops.
Below is a simple table that you can plot on graph paper. The x‑axis is volume (\$V\$) and the y‑axis is pressure (\$p\$). The points will fall on a hyperbola.
| \$V\$ (m³) | \$p\$ (kPa) |
|---|---|
| 0.5 | 200 |
| 1.0 | 100 |
| 1.5 | 66.7 |
| 2.0 | 50 |
When you plot these points, you’ll see a curve that goes steeply downwards – that’s the hyperbola showing how \$p\$ and \$V\$ are inversely related.
Imagine a water balloon (the gas). If you squeeze it, the volume decreases but the pressure inside rises – just like \$pV\$ staying the same. If you let the balloon expand, the pressure falls. This everyday experience helps you remember the inverse relationship.
Suppose a gas occupies \$V1 = 1.0\,\text{m}^3\$ at a pressure of \$p1 = 100\,\text{kPa}\$. If the volume is suddenly doubled to \$V2 = 2.0\,\text{m}^3\$, what is the new pressure \$p2\$?
??
Result: The pressure halves when the volume doubles.
Fill in the blanks: If the pressure of a gas is \$p1\$ and the volume is \$V1\$, then the pressure when the volume is doubled is .
Answer: \$p_1/2\$ (because \$pV\$ stays constant).