Question

A `4 L` container holds `4` mol and `4` mol of gasses A and B, respectively. Groups of four of molecules of gas B bind to three molecule of gas A and the reaction changes the temperature from `240^oK` to `320^oK`. By how much does the pressure change?

Answer 1

It is given that molecules of gases `Aand B` bind. Lets presume that these are diatomic gases. Monoatomic gases are inert gases and therefore may not bind with molecules of other gases. It is also presumed that there is no chemical reaction and gas molecules bind with other each other just to change the physical properties of the gases. Therefore, Dalton's law of partial pressures is applicable.

The balanced equation is

`" "3A " "+" " 4B to [(3A)(4B)]`
Initial`" "4\ mol " "4\ mol" "0\ mol"`
Final `" "1\ mol " "0\ mol" "1\ mol`

The binding action requires `3\ mol` of A for every `4\ mol` of B.

All `4` `mol` of B will bind, with `3` `mol` of A, producing `1\ mol` of `A_3B_4`. There will still be `1\ mol` of excess A in the container.
As such gas B will be the limiting participant.

We have the Ideal Gas equation as

`PV=nRT`
where `P` is the pressure of the gas, `V` is the volume of the gas, `n` is the amount of substance of gas (in moles), `R` is universal gas constant, equal to the product of the Boltzmann constant and the Avogadro constant and `T` is the absolute temperature of the gas.

Initial condition:
Using Dalton's Law of partial pressures

`P_i=(P_A+P_B)_i=((nRT_i)/V)_A+((nRT_i)/V)_B`
`=>P_i=((RT_i)/V)(n_A+n_B)`
`=>P_i=((RT_i)/V)(4+4)`
`=>P_i=8((RT_i)/V)` ........(1)

Final steady state condition. Let us call bound molecule be of Gas C. The volume does not change. The temperature of the gaseous mixture changes to `T_f`.

`P_f=(P_A+P_C)_f=((n_fRT_f)/V)_A+((nRT_f)/V)_C`
`=>P_f=((RT_f)/V)(n_(fA)+n_C)`
`=>P_f=((RT_f)/V)(1+1)`
`=>P_f=2((RT_f)/V)` ........(2)

Dividing (2) by (1) we get

`P_f/P_i=(2((RT_f)/V))/(8((RT_i)/V))`
`=>P_f/P_i=(T_f)/(4T_i)`

Inserting given temperatures we get

`P_f=(320)/(4xx240)P_i`

`=>P_f=0.bar3P_i`