# An 8 L container holds 6  mol and 15  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 310^oK to 150 ^oK. By how much does the pressure change?

Mar 6, 2018

There was a change in $57.573 \text{atm}$

#### Explanation:

We need to find $\Delta P$, and by extension, ${P}_{1}$ and ${P}_{2}$.

According to the Ideal Gas Law:

$P V = n R T$. Rearranging to solve for $P$:

$P = \frac{n R T}{V}$

Here, $V = 8 \text{L}$, ${n}_{1} = 21 \text{mol}$, ${T}_{1} = 310 \text{K}$ and $R = 0.0821 {\text{L atm K"^-1"mol}}^{-} 1$. Inputting:

${P}_{1} = \frac{21 \cdot 0.0821 \cdot 310}{8}$

${P}_{1} = 66.809 \text{atm}$

We now write down the reaction which occurred:

$6 \text{A"+15"B"rarr5"AB"_3+"A}$

So we know that ${n}_{2} = 6 \text{mol}$. Here, ${T}_{2} = 150 \text{K}$. Inputting these two new values into the Ideal Gas Law:

${P}_{2} = \frac{6 \cdot 0.0821 \cdot 150}{8}$

${P}_{2} = 9.236 \text{atm}$

We know that $\Delta P = {P}_{2} - {P}_{1}$. So here,

$\Delta P = 9.236 - 66.809 = - 57.573 \text{atm}$