# 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#