# A 7 L container holds 12  mol and 18  mol of gasses A and B, respectively. Every three of molecules of gas B bind to one molecule of gas A and the reaction changes the temperature from 350^oK to 175 ^oK. By how much does the pressure change?

Mar 9, 2016

$\setminus \Delta P = P - {P}_{0} = \left(2.494 - 12.47\right) \setminus \quad M P a = - 9.976 \setminus \quad M P a$

#### Explanation:

Ideal Gas Equation of State: $P V = n R T$
$R = 8.314 \setminus \quad J {K}^{- 1} m o {l}^{- 1}$

Before Reaction:
${P}_{A 0}$ - Partial pressure of species A;
${P}_{B 0}$ - Partial pressure of species B;
${P}_{0} = {P}_{A 0} + {P}_{B 0}$ - Total pressure before reaction;

${n}_{A 0} = 12 \setminus \quad m o l s$ - Number of moles of species A;
${n}_{B 0} = 18 \setminus \quad m o l s$ - Number of moles of species B;
${T}_{0} = 350 \setminus \quad K$ - Temperature of the mixture;
${V}_{0} = 7 \setminus \quad L i t = 7 \setminus \times {10}^{- 3} \setminus \quad {m}^{3}$ - Volume of the mixture;

P_{A0}=(n_{A0}/V)RT_0 = 4.988\quad MPa;\quad  // $1 \setminus \quad M P a = {10}^{6} \setminus \quad P a$
${P}_{B 0} = \left({n}_{B 0} / V\right) R {T}_{0} = 7.483 \setminus \quad M P a$;
${P}_{0} = {P}_{A 0} + {P}_{B 0} = 12.47 \setminus \quad M P a$

After Reaction:
$6 \setminus \quad m o l s$ of A would combine with $18 \setminus \quad m o l s$ of B to form $6 \setminus \quad m o l s$ of AB. The final mixture would be $6 \setminus \quad m o l s$ of A and $6 \setminus \quad m o l s$ of AB.

${P}_{A}$ - Partial pressure of species A;
${P}_{A B}$ - Partial pressure of the new species AB;
$P = {P}_{A} + {P}_{A B}$ - Total pressure after reaction;

${n}_{A} = 6 \setminus \quad m o l s$ - Number of moles of species A;
${n}_{A B} = 6 \setminus \quad m o l s$ - Number of moles of the new species AB;
$T = 175 \setminus \quad K$ - Temperature of the mixture after reaction;
$V = 7 \setminus \quad L i t = 7 \setminus \times {10}^{- 3} \setminus \quad {m}^{3}$ - Volume of mixture;
${P}_{A} = \left({n}_{A} / V\right) R T = 1.247 \setminus \quad M P a$;
${P}_{A B} = \left({n}_{A B} / V\right) R T = 1.247 \setminus \quad M P a$;
$P = {P}_{A} + {P}_{A B} = 2.494 \setminus \quad M P a$

Change in Pressure :
$\setminus \Delta P = P - {P}_{0} = \left(2.494 - 12.47\right) \setminus \quad M P a = - 9.976 \setminus \quad M P a$
So the pressure decreases by 9.976 MPa which in terms of percentage is a decrease by 80%