A 7 L container holds 21 mol and 12 mol of gasses A and B, respectively. Groups of three of molecules of gas B bind to five molecules of gas A and the reaction changes the temperature from 150\ K to 420\ K. How much does the pressure change?

1 Answer
Mar 8, 2018

This is what I get

Explanation:

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 gas molecules. Therefore, Dalton's law of partial pressures is applicable.

The balanced equation is

" "5A " "+" " 3B to [(5A)(3B)]
Initial 21\ mol " "12\ mol
Final " "1\ mol " "0\ mol" "1\ mol

The binding action requires 5\ mol of A for every 3\ mol of B.

All 12 mol of B will bind, with 20 mol of A, producing 1\ mol of A_5B_3. 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)(21+12)
=>P_i=33((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))/(33((RT_i)/V))
=>P_f/P_i=(2T_f)/(33T_i)

Inserting given temperatures we get

P_f=(2xx420)/(33xx150)P_i
=>P_f=0.17P_i