Question

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?

Answer 1

`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"`