A 2 L2L container holds 16 16 mol and 24 24 mol of gasses A and B, respectively. Groups of five of molecules of gas B bind to three molecules of gas A and the reaction changes the temperature from 340^oK340oK to 420^oK420oK. How much does the pressure change?

1 Answer
Jun 2, 2016

For an ideal gas, we know, (PV)/(nT) = "constant"PVnT=constant

For our case, the volume remains constant. Hence

P_1/(n_1T_1) = P_2/(n_2T_2) => P_1/P_2 = (n_1T_1)/(n_2T_2)P1n1T1=P2n2T2P1P2=n1T1n2T2

Given, T_1 = 340 KT1=340K, and T_2 = 420 KT2=420K

n_1 = 16+24 = 40n1=16+24=40

For n_2n2, note that, 16/24>3/51624>35. Hence gas B gets used up in the reaction, and in the reaction, 24/5245 moles of gas A_3B_5A3B5 is produced and (16- 24/5xx3) = 8/5(16245×3)=85 moles of gas A remains. Hence,

n_2 = 24/5+8/5 = 32/5n2=245+85=325

Hence, P_1/P_2 = (n_1T_1)/(n_2T_2) = (40xx340)/(24/5xx420) ~~6.74 P1P2=n1T1n2T2=40×340245×4206.74