An #8 L# container holds #6 # mol and #6 # mol of gasses A and B, respectively. Every two molecules of gas B bind to one molecule of gas A and the reaction raises the temperature from #320K# to #450 K#. How much does the pressure change by?

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Answer 1 · 2017-08-18T01:16:44

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Given#:# #V= 8L#; #n_1 = 6+6=12mol#

#n_2=3# mol of #A# #+3# mol of #AB_2=6# mol

#T_1=320K#; #T_2=450K#

#R=8.314J/(mol*K)#; #DeltaP=?#

Applying Ideal gas equation,

#P_1V=n_1RT_1#

#P_2V=n_2RT_2#

#=>P_2=(n_2T_2P_1)/(n_1T_1)#

#=>P_2-P_1=((n_2T_2)/(n_1T_1)-1)P_1#

#=>P_2-P_1=((n_2T_2)/(n_1T_1)-1)((n_1RT_1)/(V))#

#=>P_2-P_1=((6*450)/(12*320)-1)((12*8.314*320)/(8))#

#=>DeltaP=P_2-P_1=-1184.745J/L#

#=>DeltaP=-1184.745(Nm)/(10^-3m^3)#

#=>DeltaP=-1184745# #Pa# #=-11.84745# bar #=-11.6925# atm

#(-ve)# sign indicates that the pressure is reduced in the process.