So I learnt in class that for an ideal gas, the average kinetic energy of one atom/molecule in that gas is:

$\frac{3}{2}kT$ where k is the Boltzmann Constant ($\frac{R}{N_a}$) and T is the temperature in Kelvin. This was a derivation using a couple of other ideal gas equations.

I decided to try and use this result to work out the SHC of a gas, like hydrogen.

So I started by using the equation above to say that the difference in average kinetic energy of a single molecule in hydrogen gas when the temperature increases by 1 K is equal to $\frac{3}{2}k(T+1)-\frac{3}{2}kT = \frac{3}{2}k.$

I then considered the change in kinetic energy of an entire mole of hydrogen gas. Since the change in average kinetic energy of one single molecule is $\frac{3}{2}k$, I reasoned that the total change in kinetic energy for 1 mole of gas equals $\frac{3}{2}k$ multiplied by $N_a$ which equals $\frac{3}{2}R$.

So I have the kinetic energy required to increase the temperature of 1 mole of an ideal gas by 1 K. The mass of 1 mole of Hydrogen gas is 0.002 kg. Therefore the energy required to increase the temperature of 1 kg of hydrogen by 1 K equals $\frac{3}{2}R$ multiplied by 500. Numerically, this comes out to be 6235.5 Joules.

However, the SHC of hydrogen online is more than double this. Is there a fault in my reasoning, or is hydrogen just too far away from an ideal gas?


2 Answers 2


Some remarks before we begin. First, the quantity we want to calculate is the molar/specific heat capacity. It describes how the energy changes when you change the temperature, for a fixed amount of particles. Thus the weight of the sample is irrelevant for this calculation. Quantities that are independent of the system size are referred to as intensive quantities. The second thing I want to stress is that the changes you evaluate are differential quantities. In general, you can't just divide the change in energy for an increase of $1\rm K$ by $1\rm K$ to get the heat capacity. But rather, you should differentiate the energy function with respect to the temperature. In the special case of ideal gas the energy is linear in the temperature and thus you get the correct answer.

Now we can dive into the details. Lets make things more general. In thermodynamics there is a theorem called the equipartition theorem, which states that every degree of freedom has an average energy of $\frac{1}{2}K_{B}T$.

In your case you are talking about Hydrogen, which appears in nature in the form of the diatomic molecule $H_{2}$. This molecule has three spatial degrees of freedom (it can move in three dimensions), and as well as two rotational degrees of freedom (corresponding to the two rotation axes that are perpendicular to the molecule axis). In total there are five degrees of freedom, and thus the average energy of one molecule is


We can now directly calculate the specific heat capacity as

$$c_{V}=\frac{\partial\left<U\right>}{\partial T}=\frac{5}{2}K_{B}\approx 20.8\frac{\rm J}{\rm mole\cdot\rm K}$$

If you write "hydrogen molar heat capacity" in google you can find that the corresponding value for Hydrogen is $20.4\frac{\rm J}{\rm mole\cdot\rm K}$, pretty close to what we've just calculated.


If you are saying that the answer you got online is around double the one you calculated, it might just be due to the fact that Hydrogen gas, in equilibrium, is $H_2$ rather than $H$. So when you multiplied $E_k$, you should've doubled the quantity of moles (working with the stable $H_2$ rather than $H$). I hope this makes sense. Also, I do see that this was asked 6 years ago, but I am new into high school and was looking for a derivation of the rule and stumbled upon this. I hope this helps other curious people.

  • $\begingroup$ Welcome to Physics. I don't think that this was the OP's error; they did say that "the mass of 1 mole of hydrogen gas is 0.002 kg", and that's the mass for $\mathrm{H}_2$, not $\mathrm{H}$. $\endgroup$ Nov 21, 2022 at 12:43
  • $\begingroup$ Now you point that out, could it be the case that the online value uses a mono-atomic and idealised model of hydrogen gas? Since we are going to need to evaluate a proportionality, taking the mass of "Hydrogen gas" = 0.001 kg rather than 0.002 would mean a multiplication factor of 1000 rather than 500. $\endgroup$
    – RuhaanB
    Nov 25, 2022 at 11:32

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