# A 5 L container holds 8  mol and 10  mol of gasses A and B, respectively. Every five of molecules of gas B bind to two molecule of gas A and the reaction changes the temperature from 360^oK to 270 ^oK. By how much does the pressure change?

Feb 2, 2018

The total pressure $P$ inside the $V = 5 L$ container is due to the pressures of both gasses $A$ and $B$:

$P = {P}_{A} + {P}_{B}$.

According to the ideal gas law:

$P V = n R T$.

If we consider gasses $A$ and $B$ to be ideal, then this equation must apply to each one of them. For the total pressure, this means that:

$P = \frac{{n}_{A} R {T}_{A}}{V} _ A + \frac{{n}_{B} R {T}_{B}}{V} _ B$.

If we also assume that both gasses can occupy all the container's volume and that they will always be at thermal equilibrium with each other, then:

$P = \frac{R T}{V} \left({n}_{A} + {n}_{B}\right)$.

This means that we can obtain the pressure's value if we know the above quantities. Since $R = 8 , 2 \frac{J}{m o l K}$ is constant and we know the values for $T$ and $V$, we must figure out what happens to the number of mols of each gas after the reaction takes place.

Since it takes 5 molecules of gas $B$ to bind to 2 molecules of gas $A$, we can extend this proportion to their corresponding number of mol. Since initially we have ${n}_{1 A} = 8$ mol and ${n}_{1 B} = 10$ mol, then, after the reaction takes place, we will end up with ${n}_{2 A} = 4 m o l$, ${n}_{2 B} = 0$ mol and ${n}_{C} = 2$ mol of the new gas, the product of the $A$ and $B$ reaction.

Then, our final pressure will be described by:

${P}_{2} = \frac{R T}{V} \left({n}_{A} + {n}_{B} + {n}_{C}\right)$.

(Remember that ${P}_{1} = \frac{R T}{V} \left({n}_{A} + {n}_{B}\right)$.

Taking the values to perform the calculations:

${P}_{1} = \frac{0.082 \frac{a t m \cancel{L}}{\cancel{m o l} \cancel{K}} \cdot 360 \cancel{K}}{5 \cancel{L}} \cdot \left(8 + 10\right) \cancel{m o l}$;

${P}_{1} = 106.3 a t m$.

After the reaction takes place:

${P}_{2} = \frac{0.082 \frac{a t m \cancel{L}}{\cancel{m o l} \cancel{K}} \cdot 270 \cancel{K}}{5 \cancel{L}} \cdot \left(4 + 0 + 2\right) \cancel{m o l}$;

${P}_{2} = 26.6 a t m$.

Therefore, the change in pressure, $\Delta P$, is:

$\Delta P = {P}_{2} - {P}_{1}$;

$\Delta P = - 79 , 7 a t m$.