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.
