# 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?

Apr 13, 2017

30.377atm at 370K
34.8925atm at 425K

#### Explanation:

So first we need to know the moles of gas $A {B}_{2}$ formed

So this is chemistry stoichiometry

$2 B + A = A {B}_{2}$

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 $A {B}_{2}$

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

$\textcolor{red}{P V = n R T}$

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

$8 \times P = 8 m o l \cdot 0.0821 L \cdot {370}^{\circ} K$

$P = 0.0821 L \times {370}^{\circ} K = 30.377 a t m$

Now calculate the pressure at different temperature

$8 \times P = 8 m o l \cdot 0.0821 L \cdot {425}^{\circ} K$

$P = 0.0821 L \times {425}^{\circ} K = 34.8925 a t m$