A 5 L5L container holds 16 16 mol and 6 6 mol of gasses A and B, respectively. Every three of molecules of gas B bind to four molecule of gas A and the reaction changes the temperature from 320^oK320oK to 450 ^oK450oK. By how much does the pressure change?

1 Answer
Apr 21, 2016

56.4%56.4% increase

Explanation:

Caution : Dont write X^o Xo if it is in kelvin, write X KXK straightaway.

Considering they are ideal gases, they follow

PV = nRTPV=nRT

Here, VV is constant and RR is anyway constant, so we are left with

P_1/(n_1 T_1)=P_2/(n_2 T_2)P1n1T1=P2n2T2

Note that, 16/6>4/3166>43. So gas B gets totally consumed up. So, 6/3 = 263=2 moles of new has was formed, so in the process 88 moles of gas A was taken up and 88 moles of A is remaining. So in the reaction, total n_1 (=16+6 = 22)n1(=16+6=22) moles of gas is transformed into n_2(=2+8=10)n2(=2+8=10) moles of gas.

we are given,

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

So, P_1/P_2 = (n_1T_1)/(n_2T_2) = (22*320)/(10*450) ~~1.564 P1P2=n1T1n2T2=22320104501.564

So pressure is about 56.4%56.4% increased.