An 8 L8L container holds 6 6 mol and 15 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^oK310oK to 150 ^oK150oK. By how much does the pressure change?

1 Answer
Mar 6, 2018

There was a change in 57.573"atm"57.573atm

Explanation:

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"