An 8L 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 370oK to 425oK. How much does the pressure change by?

1 Answer
Apr 13, 2017

30.377atm at 370K
34.8925atm at 425K

Explanation:

So first we need to know the moles of gas AB2 formed

So this is chemistry stoichiometry

2B+A=AB2

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

So if 1mol of AB22mol of B thus

x mol of AB29mol of B

x = 4.5mol of AB2

Since 1 mol of AB21 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

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

8×P=8mol0.0821L370K

P=0.0821L×370K=30.377atm

Now calculate the pressure at different temperature

8×P=8mol0.0821L425K

P=0.0821L×425K=34.8925atm