Say we have liquid water. We are given specific heat of water $C=4.2kJ(kg*K)$, a number of molecules in a mol $N_A=6*10^{23}$. The atomic weight of water is $18g/mol$, and the Boltzmann's constant is $\frac{7}{5}*10^{-23}J/K$. Calculate the thermodynamic degrees of freedom of liquid water.
I am quite confused by this exercise.
Water molecules consist of one oxygen atom and two hydrogen atoms. So one water molecule obviously has 3 degrees of translational, and 2 vibrational and rotational freedom. So 3+2+2=7.
I don't understand why I was given all these useless information, or did I miss out on something. I admit Boltzmann is something i don't understand and I don't know why we would ever need Boltzmann's constant.
I just realised this is from the equipartition theorem.
where we can use: $E_{kin}=\frac{f}{2}K_B T$
So $Q=mC\Delta T=\frac{f}{2}K_B T \rightarrow f=\frac{2mC}{K_B}$
Can I take $m=1kg$ here so the units will work out? But then the exercise wouldn't give me the information on atomic weight and Avagadro's constant?
Edit: Ah I can get the mass of one water molecule by dividing the atomic weight of water by Avagadro's constant
Best Answer
The heat capacity only includes degrees of freedom that are actually accessible to the system. In a quantized system, if the first excited state has energy much larger than $kT$, then thermal interactions can’t activate that degree of freedom and it doesn’t contribute.
You have already done this implicitly by treating your water molecule as three atoms, rather than three nuclei and ten electrons. When it gets hot enough to drive electronic excitations, you tend to dissociate the molecule. You could also have asked about the nuclear degrees of freedom, which are frozen out for temperatures below a mega-eV.
A famous example is hydrogen gas, whose rotational degrees of freedom (first excited state: 15 meV) are frozen out at its triple point (30 kelvin = 2.5 meV), but accessible at room temperature (25 meV). Cold hydrogen has a heat capacity like a monoatomic gas, while warm hydrogen has a heat capacity like a rigid rotor. I used to know whether the vibrational states are accessible before hot hydrogen dissociates, but I have forgotten.
You have predicted how many degrees of freedom a isolated water molecule could have. The question is how many degrees of freedom a water molecule effectively has, in liquid, which may include some neighbor-interaction stuff.