In terms of Charles' law,explain why $-273^{\circ} C$ is the lowest possible temperature.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) Charles' law states that at constant pressure,the volume $(V)$ of a fixed mass of gas is directly proportional to its absolute temperature $(T)$: $V \propto T$ or $V = kT$.
It is observed that for all gases,the plot of volume versus temperature (in $^{\circ} C$) is a straight line. If this line is extrapolated to zero volume,it intersects the temperature axis at $-273^{\circ} C$.
Mathematically,if $V = V_0(1 + \frac{t}{273.15})$,then at $t = -273.15^{\circ} C$,the volume $V$ becomes zero.
Since a gas cannot have zero or negative volume,the temperature cannot go below $-273.15^{\circ} C$ (often rounded to $-273^{\circ} C$). In reality,all gases liquefy or solidify before reaching this temperature,making it the theoretical absolute zero.

Explore More

Similar Questions

The compressibility factor of a gas is defined as $Z = PV/RT$. The compressibility factor of an ideal gas is:

At $0^{\circ}C$,the density of a certain oxide of a gas at $2 \ bar$ is the same as that of dinitrogen at $5 \ bar$. What is the molecular mass of the oxide?

In the equation $PV = nRT$,which of the following is $NOT$ equal to the value of $R$?

Five grams each of the following gases at $87 \, ^{\circ}C$ and $750 \, mm$ pressure are taken. Which of them will have the least volume?

In the equation $PV = nRT$,the number of moles per liter can be expressed as:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo