A 4 L4L container holds 4 4 mol and 4 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^oK240oK to 320^oK320oK. By how much does the pressure change?

1 Answer
Apr 4, 2018

It is given that molecules of gases Aand BAandB 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)] 3A + 4B[(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