A 2L container holds 16 mol and 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 340oK to 420oK. How much does the pressure change?

1 Answer
Jun 2, 2016

For an ideal gas, we know, PVnT=constant

For our case, the volume remains constant. Hence

P1n1T1=P2n2T2P1P2=n1T1n2T2

Given, T1=340K, and T2=420K

n1=16+24=40

For n2, note that, 1624>35. Hence gas B gets used up in the reaction, and in the reaction, 245 moles of gas A3B5 is produced and (16245×3)=85 moles of gas A remains. Hence,

n2=245+85=325

Hence, P1P2=n1T1n2T2=40×340245×4206.74