# A 6 L container holds 4  mol and 6  mol of gasses A and B, respectively. Groups of three of molecules of gas B bind to two molecules of gas A and the reaction changes the temperature from 480^oK to 270^oK. How much does the pressure change?

Nov 12, 2017

Final Pressure : ${P}_{f} = \left(\setminus \frac{{n}_{f} {T}_{f}}{{n}_{i} {T}_{i}}\right) {P}_{i} = 0.3765$ $M P a$
Pressure Change :
$\setminus \Delta P = {P}_{f} - {P}_{i} = \left[\setminus \frac{{n}_{f} {T}_{f}}{{n}_{i} {T}_{i}} - 1\right] {P}_{i} = - 2.9706$ $M P a$

#### Explanation:

Ideal Gas Equation of State: $P V = n R T$,

The volume $V$ is held constant;
$R = 4.184 \frac{J}{m o l . K}$ 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.

$P = \left(\frac{R}{V}\right) n T$ : 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: $4 A + 6 B \setminus \rightarrow 2 {A}_{2} {B}_{3}$

$4$ mols of $A$ combine with $6$ mols of $B$ to give $2$ mols of ${A}_{2} {B}_{3}$

Before Reaction: ${P}_{i} V = \left({n}_{a} + {n}_{b}\right) R {T}_{i}$
${n}_{a} = 4$ mols; \quad n_b = 6 mols; \qquad n_i=n_a+n_b=10 $m o l s$
 T_i=480 K; \qquad V=6L=6\times10^{-3}m^3
${P}_{i} = \left({n}_{a} + {n}_{b}\right) \setminus \frac{R {T}_{i}}{V} = 3.3472 \setminus \times {10}^{6}$ $P a = 3.3472$ $M P a$.

After Reaction: ${P}_{f} V = {n}_{a b} R {T}_{f}$
${n}_{f} = {n}_{a b} = 2$ mols;
 T_f=270 K; \qquad V=6L=6\times10^{-3}m^3

${P}_{f} = \left(\setminus \frac{{n}_{f} {T}_{f}}{{n}_{i} {T}_{i}}\right) {P}_{i} = \left(\setminus \frac{2 \setminus \times 270 K}{10 \setminus \times 480 K}\right) \setminus \times 3.3472$ $M P a$
$\setminus q \quad = 0.3765$ $M P a$
$\setminus \Delta P = {P}_{f} - {P}_{i} = \left[\setminus \frac{{n}_{f} {T}_{f}}{{n}_{i} {T}_{i}} - 1\right] {P}_{i} = - 2.9706$ $M P a$

The pressure change is negative, indicating that the pressure decreases.