# A 7 L container holds 38  mol and 15  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 140^oK to 210^oK. How much does the pressure change?

Feb 21, 2017

The final pressure is 0.93 times whatever the initial pressure was.

#### Explanation:

5A + 3B → A_5B_3 With an initial ratio of 38:15 gas B is the limiting reagent. Only 15 * 3/5 = 9.0 moles of A will be combined with 15 moles of B.

Thus, the final composition in the container will be 29 moles of A and 9 moles of ${A}_{5} {B}_{3}$ for a total of 38 moles. The original number of moles was 53.

Assuming ideal gas behavior where n = PV/RT , with a constant volume , the ratio of the temperature change in ${.}^{o} K$ is the same as the ratio of the change in the total number of moles of gas in the container times the inverse ratio of the pressures.

${\left({n}_{1} \cdot \frac{P V}{T}\right)}_{1} = {\left({n}_{2} \cdot \frac{P V}{T}\right)}_{2}$ ; $\left({n}_{1} / {n}_{2}\right) \cdot \left({T}_{1} / {T}_{2}\right) = {P}_{2} / {P}_{1}$

${P}_{1} / {P}_{2} = \left({n}_{2} / {n}_{1}\right) \cdot \left({T}_{2} / {T}_{1}\right)$

${P}_{1} / {P}_{2} = \left(\frac{38}{53}\right) \cdot \frac{210}{140}$

P_1/P_2 = 1.07 ; P_2 = P_1 * 0.93
so the final pressure is 0.93 times whatever the initial pressure was.