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
This is what I get
Explanation:
It is given that molecules of gases
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
All
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
#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#