A 5L 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 320oK to 450oK. By how much does the pressure change?

1 Answer
Apr 21, 2016

56.4% increase

Explanation:

Caution : Dont write Xo if it is in kelvin, write XK straightaway.

Considering they are ideal gases, they follow

PV=nRT

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

P1n1T1=P2n2T2

Note that, 166>43. So gas B gets totally consumed up. So, 63=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 n1(=16+6=22) moles of gas is transformed into n2(=2+8=10) moles of gas.

we are given,

T1=320K
T2=450K

So, P1P2=n1T1n2T2=22320104501.564

So pressure is about 56.4% increased.