Charles' Law and Absolute Zero

15th Nov 2019 @ 3 min read

We can determine the absolute zero temperature from Charles' law. When we plot a volume vs temperature (in °C) graph, the x-intercept of the graph is the absolute zero temperature. The graph can be obtained by several experimental methods. But before jumping into that let us take a quick revision of what is absolute zero and Charles' law.

Absolute zero

The absolute zero temperature is the lowest possible temperature. At such an extreme temperature, molecular motions like vibrational, translational, rotational ceases. And its value is −273.15 °C or 0 K.

Charles' law

Charles' law is an empirical law that states when the pressure of a fixed amount of gas is constant, the volume is directly proportional to its temperature. In mathematical terms, V = kT.

Relation between both

Both are related by the graph below. It is a volume versus temperature (in °C) graph plotted at constant pressure.

Charles' law and absolute zero — The x-intercept is at t = −273.15 °C.

As per Charles' law, when the temperature of an ideal gas decreases, its volume also decreases. Volume approaches zero at t = −273.15 °C, the x-intercept of the graph. If we extrapolate the line below that temperature, it will give negative volume, which is nonsense. Thus, −273.15 °C is the lowest possible temperature.

We can also prove this using the equation of Charles' law. It is V = kT. Here, T is the temperature expressed in the kelvin. V is zero only when T = 0 K, which is −273.15 °C.

$\lim_{T \to 0} V = \lim_{T \to 0} kT = 0$

Note: Only ideal gases exist at lower temperatures. Real gases will start liquefying as temperature decreases temperatures.

If we have volume versus temperature experimental data at constant pressure, the VT graph could be generated. The graph will yield a straight line similar to above, and the x-intercept will be absolute zero.

Problem

Statement: Let say V₁, t₁ and V₂, t₂ are two points on the graph of volume versus temperature. t₀ is the absolute zero temperature (in °C), and V₀ is the volume at that temperature. Derive an expression for t₀ in terms of the rest variables?

Solution: As per Charles' law,

$\frac{V_1}{T_1}=\frac{V_2}{T_2}=\frac{V_0}{T_0}$

Here, the temperatures are in the kelvin. Converting them in the degree celsius,

$\frac{V_1}{t_1+273.15}=\frac{V_2}{t_2+273.15}=\frac{V_0}{t_0+273.15}$ $\frac{V_1}{V_2}=\frac{t_1+273.15}{t_2+273.15} \qquad \text{and} \qquad \frac{V_0}{V_2}=\frac{t_0+273.15}{t_2+273.15}$

By dividendo,

$\frac{V_1-V_2}{V_2}=\frac{t_1-t_2}{t_2+273.15} \qquad \text{and} \qquad \frac{V_0-V_2}{V_2}=\frac{t_0-t_2}{t_2+273.15}$

Taking the ratio of the former to the latter,

$\frac{V_1-V_2}{V_0-V_2}=\frac{t_1-t_2}{t_0-t_2}$

We know volume at absolute zero is zero, so V₀ = 0.

$\frac{V_1-V_2}{0-V_2}=\frac{t_1-t_2}{t_0-t_2}$ $(V_1-V_2)(t_0-t_2)=-V_2(t_1-t_2)$ $t_0V_1-t_0V_2-t_2V_1+t_2V_2=-t_1V_2+t_2V_2$ $t_0(V_1-V_2)=t_2V_1-t_1V_2$

Rearranging, we get the final expression.

$t_0=\frac{t_2V_1-t_1V_2}{V_1-V_2}$

By knowing V₁, t₁ and V₂, t₂, we can determine t₀.

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Thanks for your response!

ABDULWASAY
22nd Nov 2022

NICE ,IT WAS REALLY HELPFULL

Hritu Dey
12th May 2021

It's really easy-peasy topic