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?
1 Answer
It is given that molecules of gases
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
All
As such gas B will be the limiting participant.
We have the Ideal Gas equation as
PV=nRT
whereP 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 andT 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
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