# A 2 L container holds 16  mol and 24  mol of gasses A and B, respectively. Groups of five of molecules of gas B bind to three molecules of gas A and the reaction changes the temperature from 340^oK to 420^oK. How much does the pressure change?

Jun 2, 2016

For an ideal gas, we know, $\frac{P V}{n T} = \text{constant}$

For our case, the volume remains constant. Hence

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

Given, ${T}_{1} = 340 K$, and ${T}_{2} = 420 K$

${n}_{1} = 16 + 24 = 40$

For ${n}_{2}$, note that, $\frac{16}{24} > \frac{3}{5}$. Hence gas B gets used up in the reaction, and in the reaction, $\frac{24}{5}$ moles of gas ${A}_{3} {B}_{5}$ is produced and $\left(16 - \frac{24}{5} \times 3\right) = \frac{8}{5}$ moles of gas A remains. Hence,

${n}_{2} = \frac{24}{5} + \frac{8}{5} = \frac{32}{5}$

Hence, ${P}_{1} / {P}_{2} = \frac{{n}_{1} {T}_{1}}{{n}_{2} {T}_{2}} = \frac{40 \times 340}{\frac{24}{5} \times 420} \approx 6.74$