Ideal Gas Equation of State: #PV=nRT#
#R = 8.314\quadJK^{-1}mol^{-1}#
Before Reaction:
#P_{A0}# - Partial pressure of species A;
#P_{B0}# - Partial pressure of species B;
#P_0 = P_{A0}+P_{B0}# - Total pressure before reaction;
#n_{A0} = 12\quad mols# - Number of moles of species A;
#n_{B0}=18\quad mols# - Number of moles of species B;
#T_0=350\quad K# - Temperature of the mixture;
#V_0=7\quad Lit = 7\times10^{-3} \quad m^3# - Volume of the mixture;
#P_{A0}=(n_{A0}/V)RT_0 = 4.988\quad MPa;\quad # // #1 \quadMPa=10^6\quad Pa#
#P_{B0}=(n_{B0}/V)RT_0 = 7.483\quad MPa#;
#P_0 = P_{A0}+P_{B0} = 12.47\quad MPa#
After Reaction:
#6\quad mols# of A would combine with #18\quad mols# of B to form #6\quad mols# of AB. The final mixture would be #6\quad mols# of A and #6\quad mols# of AB.
#P_{A}# - Partial pressure of species A;
#P_{AB}# - Partial pressure of the new species AB;
#P=P_{A}+P_{AB}# - Total pressure after reaction;
#n_{A} = 6\quad mols# - Number of moles of species A;
#n_{AB}=6\quad mols# - Number of moles of the new species AB;
#T=175\quad K# - Temperature of the mixture after reaction;
#V=7\quad Lit = 7\times10^{-3} \quad m^3# - Volume of mixture;
#P_{A}=(n_{A}/V)RT = 1.247 \quad MPa#;
#P_{AB}=(n_{AB}/V)RT = 1.247\quad MPa#;
#P = P_{A}+P_{AB} = 2.494\quad MPa#
Change in Pressure :
#\Delta P = P-P_0 = (2.494-12.47)\quad MPa = -9.976\quadMPa#
So the pressure decreases by 9.976 MPa which in terms of percentage is a decrease by #80%#
