Question

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?

Answer 1

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"`