A $7 L$ $7 L$ container holds $38$ $38$ mol and $15$ $15$ 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 $140^{o} K$ $140^oK$ to $210^{o} K$ $210^oK$ . How much does the pressure change?

Question

A $7 L$ $7 L$ container holds $38$ $38$ mol and $15$ $15$ 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 $140^{o} K$ $140^oK$ to $210^{o} K$ $210^oK$ . How much does the pressure change?

1 Answer

Impact of this question

1565 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-21T02:15:45

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

Explanation:

$5 A + 3 B \to A_{5} B_{3}$ $5A + 3B → A_5B_3$ 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 $A_{5} B_{3}$ $A_5B_3$ 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 $.^{o} K$ $.^oK$ 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.

${(n_{1} \cdot \frac{P V}{T})}_{1} = {(n_{2} \cdot \frac{P V}{T})}_{2}$ $(n_1*(PV)/T)_1 = (n_2*(PV)/T)_2$ ; $(\frac{n_{1}}{n_{2}}) \cdot (\frac{T_{1}}{T_{2}}) = \frac{P_{2}}{P_{1}}$ $(n_1/n_2)*(T_1/T_2) = P_2/P_1$

$\frac{P_{1}}{P_{2}} = (\frac{n_{2}}{n_{1}}) \cdot (\frac{T_{2}}{T_{1}})$ $P_1/P_2 = (n_2/n_1)*(T_2/T_1)$

$\frac{P_{1}}{P_{2}} = (\frac{38}{53}) \cdot \frac{210}{140}$ $P_1/P_2 = (38/53)*(210)/(140)$

$\frac{P_{1}}{P_{2}} = 1.07; P_{2} = P_{1} \cdot 0.93$ $P_1/P_2 = 1.07 ; P_2 = P_1 * 0.93$
so the final pressure is 0.93 times whatever the initial pressure was.

Science

Math

Humanities

... and beyond

A $7 L$ $7 L$ container holds $38$ $38$ mol and $15$ $15$ 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 $140^{o} K$ $140^oK$ to $210^{o} K$ $210^oK$ . How much does the pressure change?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

A 7L7 L container holds 3838 mol and 1515 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 140oK140^oK to 210oK210^oK. How much does the pressure change?

1 Answer

Explanation:

Related questions

Impact of this question

A $7 L$ $7 L$ container holds $38$ $38$ mol and $15$ $15$ 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 $140^{o} K$ $140^oK$ to $210^{o} K$ $210^oK$ . How much does the pressure change?