Verification of the relationship between the volume and the temperature of a gas at constant pressure

According to the Figure 23.1, if the volume of the air trapped in the tube is $V$ and the Kelvin temperature of the gas is $T$, then according to Charles' law,

$V\propto T$ of a fixed mass of a gas at constant pressure

$$\therefore V = KT$$

If the length of the air column is $l$ and the area of internal cross-section of the tube is $A$,

$$V = lA$$ $$\therefore lA = KT$$ $$l=\frac{K}{A}T$$

If the graph of $l$ plotted against $T$ passes through the origin as a straight line, the relationship between the volume of the gas and its temperature is verified.

<image>

If $V$ is the volume of the air enclosed in the bent portion of the tube and $A$ the area of internal cross section of the tube,

$$A~l+V=KT$$ $$l=(\frac{K}{A})T-\frac{V}{A}$$

1. Attach the tube to the scale so that the closed end of the tube coincides with the zero of the scale and fix the thermometer placing its bulb close to the middle of the air column.
2. Set up the apparatus as shown in the Figure 23.1 and note down the reading of the thermometer and the length of the air column.
3. Heat the water in the beaker while being stirred. When the temperature has increased by about $10^{\circ}C$ remove the burner, stir the water and maintaining a constant reading of the thermometer note down the reading of the thermometer and the length of the air column again when the mercury index becomes still.
4. Continue the heating of the water in the beaker while being stirred and obtain six pairs of readings in this manner at every $10^{\circ}C$ rise of temperature.
5. Plot $l$ against $T$.