A 7 L container holds 30 mol and 18 mol of gasses A and B, respectively. Groups of three of molecules of gas B bind to five molecules of gas A and the reaction changes the temperature from 320^oK to 480^oK. How much does the pressure change?

1 Answer
Mar 1, 2018

DeltaP=-150.59

Explanation:

We have the Ideal Gas Law:

color(white)(ffffffff)PV=nRT, where R=0.0821"L atm/K mol"

To find DeltaP, we must find P_1 and P_2. First, we have:

P_1V_1=n_1RT_1. Here, R is the same because it is a constant.

We have to solve for P_1:

color(white)(ffffffffffffffffff)P_1=(n_1RT_1)/V_1

Here, we have:

n_1=Sigman_i=30+18=48"mol"

V_1=7"L"

T_1=320"K"

Inputting:

P_1=(48*0.0821*320)/7

P_1=180.15"atm"

We need to write down a balanced equation of the reaction:

30"A"+18"B"rarr6"A"_5"B"_3

So Sigman_i becomes 6"mol". Our new ideal gas equation is:

P_2V_2=n_2RT_2, or:

P_2=(n_2RT_2)/V_2, where:

n_2=6"mol"

V_2=7"L"

T_2=480"K"

Inputting:

P_2=(6*0.0821*420)/7

P_2=29.56"atm"

Remember that, DeltaP=P_2-P_1, so:

DeltaP=29.56-180.15

DeltaP=-150.59

So the change in pressure was 150.59"atm"