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?

Answer 1

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

Explanation:

The key to this problem is to find the number of moles of gas before and after the reaction.

We can then use the ideal gas equation to find the pressure in each case and hence the pressure change.

Before the gases react we have a total number of moles of 7 + 18 = 25

We can write the equation 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.

So the total moles after reaction = 6 + 1 = 7.

`:.` Total initial moles = 25

Total final moles = 7

The ideal gas equation gives us :

`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

Initial:

`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)`)

Final:

`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)`