Ideal Gas Equation of State: #PV=nRT#,
The volume #V# is held constant;
#R=4.184 J/(mol.K)# is the universal gas constant.
Everything else (#P#, #n# and #T#) vary. So let us rewrite the EoS making this fact explicit by grouping all the terms that are held constant inside a parenthesis.
#P=(R/V)nT# : terms inside the parenthesis are constant
(#P_i#, #n_i#, #T_i#) : Pressure, number of moles and temperature before the reaction.
(#P_f#, #n_f#, #T_f#) : Pressure, number of moles and temperature after the reaction.
#P_i = (R/V) n_iT_i; \qquad P_f = (R/V)n_fT_f; \qquad P_f/P_i = (n_fT_f)/(n_iT_i); #
#P_f = (\frac{n_fT_f}{n_iT_i})P_i; \qquad \DeltaP = P_f-P_i = [\frac{n_fT_f}{n_iT_i}-1]P_i#
Stoichiometry: #4A+6B \rightarrow 2A_2B_3#
#4# mols of #A# combine with #6# mols of #B# to give #2# mols of #A_2B_3#
Before Reaction: #P_iV=(n_a+n_b)RT_i#
#n_a = 4# #mols; \quad n_b = 6# #mols; \qquad n_i=n_a+n_b=10# #mols#
# T_i=480 K; \qquad V=6L=6\times10^{-3}m^3#
#P_i = (n_a+n_b)\frac{RT_i}{V} = 3.3472\times10^6# #Pa = 3.3472# #MPa#.
After Reaction: #P_fV = n_{ab}RT_f#
#n_f = n_{ab} = 2# #mols;#
# T_f=270 K; \qquad V=6L=6\times10^{-3}m^3#
#P_f = (\frac{n_fT_f}{n_iT_i})P_i = (\frac{2\times270K}{10\times480K})\times3.3472 # #MPa#
#\qquad= 0.3765# #MPa#
# \DeltaP = P_f-P_i = [\frac{n_fT_f}{n_iT_i}-1]P_i = -2.9706# #MPa#
The pressure change is negative, indicating that the pressure decreases.
