Question

A `5` `L` container holds `8` mol and `5` mol of gasses A and B, respectively. Groups of five molecules of gas B bind to two molecules of gas A and the reaction changes the temperature from `100` `K` to `240` `K`. How much does the pressure change?

Answer 1

The reaction decreases the number of moles of gas in the (constant volume) container from `13` to `7` while increasing the temperature from `100` `K` to `240` `K`.

The pressure change is `2794-2162=632` `kPa`

Explanation:

First we need to write a balanced chemical equation for the reaction:

`2A+5B->A_2B_5`

Since we have `5` mol of `B`, all of it will be used up in the reaction. Only `2` of the `8` mol of `A` will be used, so `6` mol will remain.

After the reaction, therefore, there will be `6` mol of `A` and `1` mol of `B` for a total of `7` mol. There were initially `13` mol of gas in the container.

We remember that a mole of any gas that behaves like an ideal gas takes up the same volume under the same temperature and pressure conditions.

Let's call the pressure before the reaction `P_1` and after the reaction the pressure is `P_2`.

`PV=nRT` where `R` is the gas constant, `8.314` `LkPaK^-1mol^-1`

`P_1=(nRT)/V=(13xx8.314xx100)/5=2162` `kPa`

`P_2=(nRT)/V=(7xx8.314xx240)/5=2794` `kPa`

The pressure change is `2794-2162=632` `kPa`

Answer 2

29% increase

Explanation:

Idea gas law:

PV=nRT

For mix gases inside the same volume and temperature

PV`= (n_A+n_B)RT`

Let `(n_1, P_1, V_1, T_1) and (n_2, P_2, V_2, T_1)` denote the number of moles, pressure, volume, and temperature before and after the reaction happened respectively.

Before the reaction, there are

`n_1= n_A+n_B= 8 + 5= 13 mol` of mixed gases

The reaction

`cancel(5B -> 2A)` (This reaction is incorrect, I misread the question, B binds with A, not binds to form A)

`5B + 2A -> A_2B_5`

Takes away 5 moles of B and 2 moles of A into one moles of `A_2B_5`, leaving 6 mol of A.

`n_2= n_A+n_(A_2B_5)` = 6+1 = 7 mol of mixed gases.

The idea gas law before and after the reaction are:

`P_1V_1 = n_1 RT_1`

`P_2V_2 = n_2RT_2`

`(P_2V_2)/(P_1V_1) = n_2/n_1(RT_2)/(RT_1)`

For a fixed container, `V_1 = V_2`

`P_2/(P_1) = 7/13 * (240K)/(100K) = 1.29`

The change in pressure is

`(P_2-P_1)/p_1 = (DeltaP)/P_1 =P_2/P_1 -1 = 0.29 or 29%`

The pressure has thus increased 85%.

Since the volume of the container is given, `P_1` can be reality calculated to determine the absolute change in pressure (`DeltaP`)

`P_1 = (n_1 RT_1)/V = ((13mol*8.3145 J/(mol K)*100K)/(5L*0.001m^3/L)) /(1.01325 "x" 10^5 N/(m^2atm)) ~ 21.34 atm`

`DeltaP ~0.29P_1= 0.29* 21.3 atm = 6.2 atm`