A #5 L# container holds #5 # mol and #10 # mol of gasses A and B, respectively. Every five of molecules of gas B bind to two 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
Feb 24, 2018

There is a change of #-56.649# atmospheres.

Explanation:

To solve for #DeltaP#, we need two values: #P_"initial"# and #P_"final"#.

First, we calculate the initial pressure using the Ideal Gas Law:

#P_iV=n_iRT_i#

Here,

#V=5"L"#

#n_i=15"mol"#

#T_i=320"K"#

#R=0.0821"L atm K"^-1"mol"^_1#

To solve for #P_i#, we rearrange:

#P_i=(n_iRT_i)/V#

And input:

#P_i=(15*0.0821*320)/5#

#P_i=78.816"atm"#

Now, we need the equation that took place. It is:

#5"A"+10"B"rarr2"A"_2"B"_5+"A"#. Here, one mole of #"A"# is left unreacted.

In the products side, we find we now have #3# moles of gas. Use the Ideal Gas Law again:

#P_fV=n_fRT_f#

Here,

#n_f=3"mol"#

#T_f=450"K"#

Rearranging to solve for #P_f#:

#P_f=(n_fRT_f)/V# and inputting:

#P_f=(3*0.0821*450)/5#

#P_f=22.167"atm"#

We know that #DeltaP=P_f-P_i#. Inputting:

#DeltaP=22.167-78.816#

#DeltaP=-56.549"atm"#