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.
