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"#