# A 9 L container holds 8  mol and 2  mol of gasses A and B, respectively. Every two molecules of gas B bind to one molecule of gas A and the reaction raises the temperature from 370^oK to 425 ^oK. How much does the pressure change by?

Nov 29, 2017

${P}_{i} = \left({n}_{a} + {n}_{b}\right) \setminus \frac{R {T}_{i}}{V} = 1.7201$ $M P a$.
${P}_{f} = \left({n}_{a b} + {n}_{c}\right) \setminus \frac{R {T}_{f}}{V} = 1.5806$ $M P a$

$\setminus \Delta P = {P}_{f} - {P}_{i} = - 0.1395$ $M P a$

#### 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: 8 moles of $A$ and 2 moles of $B$ combine to give 1 mole of $A {B}_{2}$ and 7 moles of $A$.

$8 A + 2 B \setminus \rightarrow 7 A + A {B}_{2}$

Before Reaction: ${P}_{i} V = \left({n}_{a} + {n}_{b}\right) R {T}_{i}$
${n}_{a} = 8$ mols; \quad n_b = 2 mols; \qquad n_i=n_a+n_b=10 $m o l s$
 T_i=370 K; \qquad V= 9L = 9\times10^{-3}m^3
${P}_{i} = \left({n}_{a} + {n}_{b}\right) \setminus \frac{R {T}_{i}}{V} = 1.7201 \setminus \times {10}^{6}$ $P a = 1.7201$ $M P a$.

After Reaction: ${P}_{f} V = {n}_{f} R {T}_{f}$
${n}_{a b} = 1$ mols; \quad n_c = 7 mols; \qquad n_f = n_{ab} + n_c = 8 $m o l s$
 T_f=425 K; \qquad V = 9L = 9\times10^{-3}m^3

Because $R$ and $V$ are constants,
${P}_{f} = \left(\setminus \frac{{n}_{f} {T}_{f}}{{n}_{i} {T}_{i}}\right) {P}_{i} = \left(\setminus \frac{8 \setminus \times 425 K}{10 \setminus \times 370 K}\right) \setminus \times 1.7201$ $M P a$
$\setminus q \quad = 1.5806$ $M P a$
$\setminus \Delta P = {P}_{f} - {P}_{i} = \left[\setminus \frac{{n}_{f} {T}_{f}}{{n}_{i} {T}_{i}} - 1\right] {P}_{i} = - 0.1395$ $M P a$

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