# A 7 L container holds 30  mol and 18  mol of gasses A and B, respectively. Groups of three of molecules of gas B bind to five molecules of gas A and the reaction changes the temperature from 320^oK to 480^oK. How much does the pressure change?

Mar 1, 2018

$\Delta P = - 150.59$

#### Explanation:

We have the Ideal Gas Law:

$\textcolor{w h i t e}{f f f f f f f f} P V = n R T$, where $R = 0.0821 \text{L atm/K mol}$

To find $\Delta P$, we must find ${P}_{1}$ and ${P}_{2}$. First, we have:

${P}_{1} {V}_{1} = {n}_{1} R {T}_{1}$. Here, $R$ is the same because it is a constant.

We have to solve for ${P}_{1}$:

$\textcolor{w h i t e}{f f f f f f f f f f f f f f f f f f} {P}_{1} = \frac{{n}_{1} R {T}_{1}}{V} _ 1$

Here, we have:

${n}_{1} = \Sigma {n}_{i} = 30 + 18 = 48 \text{mol}$

${V}_{1} = 7 \text{L}$

${T}_{1} = 320 \text{K}$

Inputting:

${P}_{1} = \frac{48 \cdot 0.0821 \cdot 320}{7}$

${P}_{1} = 180.15 \text{atm}$

We need to write down a balanced equation of the reaction:

$30 {\text{A"+18"B"rarr6"A"_5"B}}_{3}$

So $\Sigma {n}_{i}$ becomes $6 \text{mol}$. Our new ideal gas equation is:

${P}_{2} {V}_{2} = {n}_{2} R {T}_{2}$, or:

${P}_{2} = \frac{{n}_{2} R {T}_{2}}{V} _ 2$, where:

${n}_{2} = 6 \text{mol}$

${V}_{2} = 7 \text{L}$

${T}_{2} = 480 \text{K}$

Inputting:

${P}_{2} = \frac{6 \cdot 0.0821 \cdot 420}{7}$

${P}_{2} = 29.56 \text{atm}$

Remember that, $\Delta P = {P}_{2} - {P}_{1}$, so:

$\Delta P = 29.56 - 180.15$

$\Delta P = - 150.59$

So the change in pressure was $150.59 \text{atm}$