Ideal Gases — Question 8

Question 8

Pressure can reveal dissociation when temperature is known

Treat the gas as ideal, $pV=nRT$, and use $R=8.314\,\mathrm{J/(mol\,K)}$. A rigid container initially contains only diatomic molecules $X_2$ at pressure 100 kPa and temperature 300 K. It is heated to 600 K, and the measured pressure is 280 kPa. A fraction $f$ of the original molecules has dissociated into two atoms, with no loss of nuclei.

1. Infer $f$ and the final mole fractions of free atoms and remaining molecules.

2. Find the pressure range consistent with partial or complete dissociation at 600 K. Explain why a pressure of 450 kPa would reject at least one assumption.

Question 8 - Solution

Pressure can reveal dissociation when temperature is known

(a) Method and calculation

$$n_f=n_i(1+f),\quad\frac{p_f}{p_i}=(1+f)\frac{T_f}{T_i}.$$$$2.80=2(1+f)\ \Rightarrow\ f=.400.$$$$x_X=\frac{2f}{1+f}=\frac47,\quad x_{X_2}=\frac{1-f}{1+f}=\frac37.$$

(b) Interpretation and further calculation

$$0\le f\le1\ \Rightarrow\ 200\le p_f\le400\,\mathrm{kPa}.$$$$p_f=450\,\mathrm{kPa}\ \Rightarrow\ f=450/200-1=1.25.$$That impossible fraction indicates an incorrect temperature, extra particles, changing volume or failure of the assumed chemical or gas model.

Physical check

Heating and increased particle number both raise pressure; they must be separated in the inference.