Question

An `8 L` container holds `8` mol and `9` mol of gasses A and B, respectively. Every two molecules of gas B bind to one molecule of gas A and the reaction raises the temperature from `370^oK` to `425 ^oK`. How much does the pressure change by?

Answer 1

30.377atm at 370K
34.8925atm at 425K

Explanation:

So first we need to know the moles of gas `AB_2` formed

So this is chemistry stoichiometry

`2B + A = AB_2`

So the reagent in excess is A and some amount of it will remain till the end

So if `("1mol of "AB_2)/"2mol of B"` thus

`("x mol of "AB_2)/"9mol of B"`

x = 4.5mol of `AB_2`

Since `("1 mol of "AB_2)/"1 mol of A"`

So the amount of AB2 is equal to the amount of A used

So A used is 4.5mol

But left out is 8 - 4.5 = 3.5mol

So mol of A + mol of AB2 =

4.5mol + 3.5mol = 8mol

Now calculate the volume of gas at 370K and 425K

using the formula

`color(red)(PV = nRT)`

Where P is pressure in atm
n is moles
R is the universal gas constant which is equal to 0.0821L
T is temperature in kelvin
V is volume

Plug in the variables

`8xxP = 8mol * 0.0821L * 370^@K`

`P = 0.0821L xx370^@K = 30.377atm`

Now calculate the pressure at different temperature

`8xxP = 8mol * 0.0821L * 425^@K`

`P = 0.0821L xx425^@K = 34.8925atm`