A 5 L container holds 16 mol and 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^oK to 450 ^oK. By how much does the pressure change?

1 Answer
Apr 21, 2016

56.4% increase

Explanation:

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

Considering they are ideal gases, they follow

PV = nRT

Here, V is constant and R is anyway constant, so we are left with

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

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

we are given,

T_1 = 320 K
T_2 = 450K

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

So pressure is about 56.4% increased.