A 5 L container holds 4 mol and 4 mol of gasses A and B, respectively. Groups of four of molecules of gas B bind to three molecule of gas A and the reaction changes the temperature from 280^oK to 320^oK. By how much does the pressure change?

1 Answer
Apr 18, 2018

A decrease of ~~2700\ "Pa"

Explanation:

We know that PV=nRT, where

  • P is pressure
  • V is volume
  • n is number of moles
  • R is the gas constant, and equal to 8.31\ "J"/("mol"\ "K")
  • T is temperature

Rearrange the formula to get P=(nRT)/V

Initially, the pressure P=((4+4)*8.31*280)/5=3722.88\ "Pa".

With the reaction, every 4\ "mol" of gas B bonds to 3\ "mol" of gas A. Since there are 4\ "mol" initially for both chemicals, there will be 1\ "mol" of gas A left and 1\ "mol" of the compound of A and B left, or 2\ "mol" of gas in total.

(You can see more clearly if you write as an equation 3A+4B->A_3B_4. Seven moles of gas become 1 mole after the reaction.)

The temperature rises to 320\ "K", while the volume stays constant.

Thus, the new pressure is P=(2*8.31*320)/5=1063.68\ "Pa".

Thus, there is a decrease in pressure of 3722.88-1063.68=2659.2~~2700\ "Pa".