# An 8 L container holds 6  mol and 6  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 320K to 450 K. How much does the pressure change by?

Aug 18, 2017

see below

#### Explanation:

Given$:$ $V = 8 L$; ${n}_{1} = 6 + 6 = 12 m o l$

${n}_{2} = 3$ mol of $A$ $+ 3$ mol of $A {B}_{2} = 6$ mol

${T}_{1} = 320 K$; ${T}_{2} = 450 K$

$R = 8.314 \frac{J}{m o l \cdot K}$; DeltaP=?

Applying Ideal gas equation,

${P}_{1} V = {n}_{1} R {T}_{1}$

${P}_{2} V = {n}_{2} R {T}_{2}$

$\implies {P}_{2} = \frac{{n}_{2} {T}_{2} {P}_{1}}{{n}_{1} {T}_{1}}$

$\implies {P}_{2} - {P}_{1} = \left(\frac{{n}_{2} {T}_{2}}{{n}_{1} {T}_{1}} - 1\right) {P}_{1}$

$\implies {P}_{2} - {P}_{1} = \left(\frac{{n}_{2} {T}_{2}}{{n}_{1} {T}_{1}} - 1\right) \left(\frac{{n}_{1} R {T}_{1}}{V}\right)$

$\implies {P}_{2} - {P}_{1} = \left(\frac{6 \cdot 450}{12 \cdot 320} - 1\right) \left(\frac{12 \cdot 8.314 \cdot 320}{8}\right)$

$\implies \Delta P = {P}_{2} - {P}_{1} = - 1184.745 \frac{J}{L}$

$\implies \Delta P = - 1184.745 \frac{N m}{{10}^{-} 3 {m}^{3}}$

$\implies \Delta P = - 1184745$ $P a$ $= - 11.84745$ bar $= - 11.6925$ atm

$\left(- v e\right)$ sign indicates that the pressure is reduced in the process.