Question

A `5 L` container holds `16` mol and `6` mol of gasses A and B, respectively. Every three of molecules of gas B bind to four molecule of gas A and the reaction changes the temperature from `320^oK` to `450 ^oK`. By how much does the pressure change?

Answer 1

`56.4%` increase

Explanation:

Caution : Dont write `X^o` if it is in kelvin, write `X K` straightaway.

Considering they are ideal gases, they follow

`PV = nRT`

Here, `V` is constant and `R` is anyway constant, so we are left with

`P_1/(n_1 T_1)=P_2/(n_2 T_2)`

Note that, `16/6>4/3`. So gas B gets totally consumed up. So, `6/3 = 2` moles of new has was formed, so in the process `8` moles of gas A was taken up and `8` moles of A is remaining. So in the reaction, total `n_1 (=16+6 = 22)` moles of gas is transformed into `n_2(=2+8=10)` moles of gas.

we are given,

`T_1 = 320 K`
`T_2 = 450K`

So, `P_1/P_2 = (n_1T_1)/(n_2T_2) = (22*320)/(10*450) ~~1.564`

So pressure is about `56.4%` increased.