A $2 L$ $2 L$ container holds $16$ $16$ mol and $24$ $24$ 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 $340^{o} K$ $340^oK$ to $420^{o} K$ $420^oK$ . How much does the pressure change?

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A $2 L$ $2 L$ container holds $16$ $16$ mol and $24$ $24$ 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 $340^{o} K$ $340^oK$ to $420^{o} K$ $420^oK$ . How much does the pressure change?

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Answer 1 · 2016-06-02T12:33:53

For an ideal gas, we know, $\frac{P V}{n T} = constant$ $(PV)/(nT) = "constant"$

For our case, the volume remains constant. Hence

$\frac{P_{1}}{n_{1} T_{1}} = \frac{P_{2}}{n_{2} T_{2}} \Rightarrow \frac{P_{1}}{P_{2}} = \frac{n_{1} T_{1}}{n_{2} T_{2}}$ $P_1/(n_1T_1) = P_2/(n_2T_2) => P_1/P_2 = (n_1T_1)/(n_2T_2)$

Given, $T_{1} = 340 K$ $T_1 = 340 K$ , and $T_{2} = 420 K$ $T_2 = 420 K$

$n_{1} = 16 + 24 = 40$ $n_1 = 16+24 = 40$

For $n_{2}$ $n_2$ , note that, $\frac{16}{24} > \frac{3}{5}$ $16/24>3/5$ . Hence gas B gets used up in the reaction, and in the reaction, $\frac{24}{5}$ $24/5$ moles of gas $A_{3} B_{5}$ $A_3B_5$ is produced and $(16 - \frac{24}{5} \times 3) = \frac{8}{5}$ $(16- 24/5xx3) = 8/5$ moles of gas A remains. Hence,

$n_{2} = \frac{24}{5} + \frac{8}{5} = \frac{32}{5}$ $n_2 = 24/5+8/5 = 32/5$

Hence, $\frac{P_{1}}{P_{2}} = \frac{n_{1} T_{1}}{n_{2} T_{2}} = \frac{40 \times 340}{\frac{24}{5} \times 420} \approx 6.74$ $P_1/P_2 = (n_1T_1)/(n_2T_2) = (40xx340)/(24/5xx420) ~~6.74$

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A $2 L$ $2 L$ container holds $16$ $16$ mol and $24$ $24$ 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 $340^{o} K$ $340^oK$ to $420^{o} K$ $420^oK$ . How much does the pressure change?

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A 2 L2L2 L container holds 16 1616 mol and 24 2424 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^oK340oK340^oK to 420^oK420oK420^oK. How much does the pressure change?

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A $2 L$ $2 L$ container holds $16$ $16$ mol and $24$ $24$ 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 $340^{o} K$ $340^oK$ to $420^{o} K$ $420^oK$ . How much does the pressure change?