While studying work done in isothermal processes I was told that the universal gas constant $R$ is the work done by the gas per mole per kelvin. I am sure I am missing something basic but I didn't find anything beyond that statement when I looked for an explanation.

This is what I found on Wikipedia:

The physical significance of $R$ is work per degree per mole.

In doing what, is 8.314 joules of work done by one mole of an ideal gas?


Charles' law says that at constant pressure the volume and temperature of an ideal gas are related as $$\frac{V_1}{T_1}=\frac{V_2}{T_2}$$ If $V_2=V_1+dV$ and $T_2=T_1+dT$ then $$\frac{V_1}{T_1}=\frac{V_1+dV}{T_1+dT}=\frac{V_1}{T_1}\left(\frac{1+dV/V_1}{1+dT/T_1}\right)$$ which can be rearranged into $$\frac{dV}{dT}=\frac{V_1}{T_1}$$ But Charles' law says that the rhs in the previous expression is a constant ratio at constant pressure and amount of substance. And by multiplying both sides by $-p$ we have that $$-p\frac{dV}{dT}=\frac{dw}{dT}=-p\frac{V_1}{T_1}$$ Then invoking the ideal gas law for one mole of substance we obtain $$\left(\frac{\partial w}{\partial T}\right)_p=-R$$ This result could of course have been obtained without using such a roundabout way, by direct differentiation of the ideal gas law (holding composition and pressure constant) or by simple finite differences (again applying Charles' law): $$w=-p\Delta V = -nR\Delta T$$

So $R$ is indeed a proportionality constant between work and temperature (a measure of thermal energy) change per mole of ideal gas.

In doing what is 8.314 joules of work done by one mole of an ideal gas?

Assume that during an expansion against constant pressure one mole of an ideal gas does an amount of work equal to $-R\cdot \pu{1 K}\cdot\pu{1 mol} = \pu{-8.3145 J}$. That is the amount of work that is done while the temperature of the gas increases by 1 kelvin (due to the heat that would need to be supplied to maintain constant pressure).

  • 1
    $\begingroup$ Yeah, that's a lot better than my answer. I don't think it's worth keeping mine around, luckily OP unaccepted it. $\endgroup$ – orthocresol May 14 at 12:56
  • $\begingroup$ @orthocresol My answer is perhaps not the best way to think of R, but it's one. The question was surprisingly tricky to get a grip on. $\endgroup$ – Buck Thorn May 14 at 13:10
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    $\begingroup$ As it turns out, thermodynamics is hard. :) $\endgroup$ – Zhe May 14 at 13:32

It may be helpful to look at a related value $k_\mathrm{B}$, the Boltzmann constant, which is widely used in thermodynamics.

These two are related by $R = k_\mathrm{B}N_\mathrm{A}$, allowing the ideal gas law to also be written: $$PV = Nk_\mathrm{B}T$$ where $N$ is the number of particles, as opposed to the number of moles.

The units are $\pu{J\cdot K^{-1}}$.

It's a proportionality between energy and temperature. In the ideal gas law, it relates temperature on the right to the product of pressure and volume, which is an energy.

This general idea is frequently used in thermodynamics, as you will see factors of the form $\exp{(-E/k_\mathrm{B}T)}$, where the $k_\mathrm{B}$ allows the exponent here to be unitless.

As examples:

  1. Planck's law, where the energy is in the form of quantum energy level spacing: $$B_{\nu}(\nu, T) = \frac{2h\nu^{3}}{c^{2}}\frac{1}{\exp\left(\frac{h\nu}{k_{\mathrm{B}}T}\right) - 1}$$
  2. Maxwell-Boltzmann distribution, where the energy refers to the kinetic energy of gas molecules: $$f(v) = \left(\frac{m}{2\pi k_{\mathrm{B}}T}\right)^{\frac{3}{2}}4\pi v^{2}\exp\left(-\frac{mv^{2}}{2k_{\mathrm{B}}T}\right)$$

$R$ is a proportionality constant between the units we use to measure temperature (kelvins) and the units we use to measure energy (joules) and number of particles (moles). If you measure temperature in energy units and use the number of particles instead of the number of moles, the constant $R$ (and $k_\mathrm{B}$) disappears from your formulas.

$k_\mathrm{B}$ and $R$ exist for historical reasons, due to choices made in the definition of the Kelvin temperature scale and the Celsius temperature scale before it, as well as the fact that the link between temperature and energy was not understood at the time when the scales were conceived. The concept of temperature seems to predate that of energy, with attempts systematically at standardizing its measurement dating to 170 CE. The modern concept of energy didn't exist until Newton's mechanics (although Aristotle had a qualitative idea of potentiality). The earliest energy concept was the vis viva, which is proportional to kinetic energy and was proposed by Leibniz around 1676. The Fahrenheit scale, the first popular quantification of temperature, was invented around 1724, and was based on the boiling point of pure water and the freezing point of a salt water mixture. A scale similar to the modern Celsius scale based on the freezing and boiling points of pure water, was proposed in 1743. The fact that mechanical energy could be converted into a change in temperature was not understood mathematically until the experiments of Joule in around 1843.

The value of $R$ was deduced from studies of mechanical properties of gases and was known at least by 1856. Its development required the concept that gases were made of small particles and the relation between the number of particles and molecular mass. The first value of Avogadro's number was estimated by Lochschmidt in 1865. This is allowed the Boltzmann constant ($k_\mathrm{B}$) to be defined, linking temperature directly to the microscopic kinetic energy of particles in an ideal gas.

The modern understanding of the relation between temperature, energy, and entropy wasn't arrived at until the work of Boltzmann and Gibbs in the late 1800s:

$$\frac{1}{k_\mathrm{B} T} = \left( \frac{\partial \ln \Omega}{\delta U} \right)_{V,N}$$

Here $U$ is the internal energy, $\Omega$ is the number of microstates corresponding to the system's macrostate, and the derivative is taken with the volume of the system ($V$) and number of particles ($N$) fixed.

Some good statistical mechanics textbooks (for example, Landau and Lifshitz), present formulas with temperature in energy units (or equivalently, with energy in temperature units), so neither $R$ nor $k_\mathrm{B}$ appear in their formulas.


From kinetic theory the total kinetic energy of an ideal gas is $\frac{1}{2}mC^2 $ where $C$ is the root mean square velocity. This is also equal to $\frac{3}{2}RT$ which means that the gas constant $R$ is equal to two thirds of the total translational kinetic energy of a mole of ideal gas at $1$ degree kelvin.


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