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?

1 Answer
Chitram Banerjee
Jun 2, 2016

For an ideal gas, we know, (PV)/(nT) = "constant"

For our case, the volume remains constant. Hence

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, and T_2 = 420 K

n_1 = 16+24 = 40

For n_2, note that, 16/24>3/5. Hence gas B gets used up in the reaction, and in the reaction, 24/5 moles of gas A_3B_5 is produced and (16- 24/5xx3) = 8/5 moles of gas A remains. Hence,

n_2 = 24/5+8/5 = 32/5

Hence, P_1/P_2 = (n_1T_1)/(n_2T_2) = (40xx340)/(24/5xx420) ~~6.74

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