A #5 L# container holds #12 # mol and #9 # 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 #360^oK# to #420 ^oK#. By how much does the pressure change?

Question

A #5 L# container holds #12 # mol and #9 # 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 #360^oK# to #420  ^oK#. By how much does the pressure change?

William Audy · Accepted Answer

the pressure change from 126 Bar to 12 Bar

the reaction is #3 B + 4 A =B_3A_4 + Q# so if at the beginning you have (12+9) = 21 mol of gases, after the reaction you have 3 mol of gas( we don't know if we have still gas?) at the beginning with the gases'law, we have: #P= nRT/V= 21 mol xx 8,31 J/(mol xx K) xx (360K)/(0,005m^3)=126 xx 10^5 J/m^3 = 126 xx 10^5 N/m^2 = 126 xx 10^5 Pascal = 126 Bar# after the reaction: #P= nRT/V= 3 mol xx 8,31 J/(mol xx K) xx (420K)/(0,005m^3)=12 xx 10^5 J/m^3 = 12 xx 10^5 N/m^2 = 12 xx 10^5 Pascal = 12 Bar#

Emily Abbate · Answer

undefined To solve this problem, we can use the ideal gas law and the concept of stoichiometry. First, we calculate the initial pressure using the ideal gas law, then we use the stoichiometry of the reaction to determine the final number of moles of gas and finally use the ideal gas law again to calculate the final pressure. The change in pressure is the difference between the initial and final pressures.

Initial pressure (P1) = (nA + nB) * R * T1 / V

Final moles of A = nA - 4 * (number of moles of B)
Final moles of B = nB - 3/4 * (number of moles of B)
Final pressure (P2) = (final moles of A + final moles of B) * R * T2 / V

Change in pressure = P2 - P1