A #7 L# container holds #7 # mol and #18 # 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 #320^oK# to #240 ^oK#. By how much does the pressure change?

Question

A #7 L# container holds #7 # mol and #18 # 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 #320^oK# to #240  ^oK#. By how much does the pressure change?

Julia Aguilar · Accepted Answer

The pressure decreases by #sf(1462color(white)(x)kPa)#

Determine the amount of gas in moles both before and after the reaction is the key to solving this puzzle.

The pressure in each situation and, consequently, the pressure change can then be determined using the ideal gas equation.

The total number of moles we have before the gases react is 7 + 18 = 25.

The equation can be expressed as:

#sf(3B_((g))+A_((g))rarrAB_(3(g))#

This tells us that 1 mole of #sf(A)# reacts with 3 moles of #sf(B)# to give 1 mole of #sf(AB_3)#.

Multiplying through by 6 tells us that 18 moles of #sf(B)# will react with 6 moles of #sf(A)# to give 6 moles of #sf(AB_3)#.

Since we have 7 moles of #sf(A)# we have 7 - 6 = 1 mole of #sf(A)# which is in excess.

Thus, after the reaction, the total moles are 6 + 1 = 7.

#:.# Total initial moles = 25

Seven final moles in total.

Using the ideal gas equation, we obtain:

#sf(PV=nRT)#

#sf(P)# is the pressure

#sf(V)# is the volume

#sf(n)# is the number of moles

#sf(R)# is the gas constant with the value #sf(8.31color(white)(x)"J/K/mol")#

#sf(T)# is the absolute temperature

First:

#sf(P_1V=n_1RT_1)#

#:.##sf(P_1=(n_1RT_1)/V)#

#sf(P_1=(25xx8.31xx320)/(0.007)=1.662xx10^(7)color(white)(x)"N/m"^2)#

(note I have converted #sf("L")# to #sf("m"^3)#)

Lastly:

#sf(P_2=(n_2RT_2)/V)#

#sf(P_2=(7xx8.31xx240)/(0.007)=0.19944xx10^(7)color(white)(x)"N""/""m"^2)#

#:.##sf(DeltaP=(1.662-0.19944)xx10^(7)=1.462xx10^(7)color(white)(x)"N""/""m"^2)#

So the pressure has been reduced by #sf(1462color(white)(x)kPa)#

Layla Ahmed · Answer

undefined To find the change in pressure, we can use the combined gas law, which states:

P1V1/T1 = P2V2/T2

Where:
P1 = initial pressure
V1 = initial volume
T1 = initial temperature
P2 = final pressure
V2 = final volume
T2 = final temperature

Given:
Initial volume (V1) = 7 L
Initial moles of gas A (nA) = 7 mol
Initial moles of gas B (nB) = 18 mol
Initial temperature (T1) = 320 K
Final temperature (T2) = 240 K

Using the ideal gas law, we can find the initial pressure (P1) and final pressure (P2):

P1 = (nA + nB) * R * T1 / V1
P2 = (nA + nB) * R * T2 / V1

Where:
R = ideal gas constant (8.314 J/(mol*K))

Substitute the given values into the equations to find P1 and P2, then calculate the difference between the final and initial pressures:

ΔP = P2 - P1