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?

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1 Answer
NickTheTurtle
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"#.

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