An #8 L# container holds #6 # mol and #15 # mol of gasses A and B, respectively. Every three of molecules of gas B bind to one molecule of gas A and the reaction changes the temperature from #310^oK# to #150 ^oK#. By how much does the pressure change?

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Answer 1 · 2018-03-06T13:00:57

There was a change in #57.573"atm"#

We need to find #DeltaP#, and by extension, #P_1# and #P_2#.

According to the Ideal Gas Law:

#PV=nRT#. Rearranging to solve for #P#:

#P=(nRT)/V#

Here, #V=8"L"#, #n_1=21"mol"#, #T_1=310"K"# and #R=0.0821"L atm K"^-1"mol"^-1#. Inputting:

#P_1=(21*0.0821*310)/8#

#P_1=66.809"atm"#

We now write down the reaction which occurred:

#6"A"+15"B"rarr5"AB"_3+"A"#

So we know that #n_2=6"mol"#. Here, #T_2=150"K"#. Inputting these two new values into the Ideal Gas Law:

#P_2=(6*0.0821*150)/8#

#P_2=9.236"atm"#

We know that #DeltaP=P_2-P_1#. So here,

#DeltaP=9.236-66.809=-57.573"atm"#