A 7 L7L container holds 21 21 mol and 12 12 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 350^oK350oK to 420^oK420oK. How much does the pressure change?
1 Answer
Let us assume that both gases are ideal gases, so we may use the ideal gas law.
Explanation:
Before reaction, we have:
V_1 = 7 " L"V1=7 L n_"A1" = 21 " mol"nA1=21 mol ,n_"B1" = 12 " mol"nB1=12 mol T_1 = 350 " K"T1=350 K
Applying the law:
Where we have used that
Now, let us write the reaction: 3 molecules of
With our initial conditions, taking in account that we need
- We would need, for 21 mol of
"A"A ,3/5 cdot 21 = 12.635⋅21=12.6 mol of"B"B . - We would need, for 12 mol of
"B"B ,5/3 cdot 12 = 2053⋅12=20 mol of"A"A .
Given that we just have 12 mol of
Thus, reaction is:
So we'll have in the end:
By applying ideal gas law again:
And the change on pressure is: