An 8 L8L container holds 6 6 mol and 6 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 320K320K to 450 K450K. How much does the pressure change by?

1 Answer
Aug 18, 2017

see below

Explanation:

Given:: V= 8LV=8L; n_1 = 6+6=12moln1=6+6=12mol

n_2=3n2=3 mol of AA +3+3 mol of AB_2=6AB2=6 mol

T_1=320KT1=320K; T_2=450KT2=450K

R=8.314J/(mol*K)R=8.314JmolK; 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.