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: 8 moles of #A# and 2 moles of #B# combine to give 1 mole of #AB_2# and 7 moles of #A#.
#8A+2B \rightarrow 7A + AB_2#
Before Reaction: #P_iV=(n_a+n_b)RT_i#
#n_a = 8# #mols; \quad n_b = 2# #mols; \qquad n_i=n_a+n_b=10# #mols#
# T_i=370 K; \qquad V= 9L = 9\times10^{-3}m^3#
#P_i = (n_a+n_b)\frac{RT_i}{V} = 1.7201\times10^6# #Pa = 1.7201# #MPa#.
After Reaction: #P_fV = n_f RT_f#
#n_{ab} = 1# #mols; \quad n_c = 7# #mols; \qquad n_f = n_{ab} + n_c = 8# #mols#
# T_f=425 K; \qquad V = 9L = 9\times10^{-3}m^3#
Because #R# and #V# are constants,
#P_f = (\frac{n_fT_f}{n_iT_i})P_i = (\frac{8\times425K}{10\times370K})\times1.7201 # #MPa#
#\qquad= 1.5806# #MPa#
# \DeltaP = P_f-P_i = [\frac{n_fT_f}{n_iT_i}-1]P_i = -0.1395# #MPa#
The pressure change is negative, indicating that the pressure decreases.
