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.