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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?

Answer 1

Charles Aeschliman

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

In our instance, the volume doesn't change.

#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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Answer 2

Adrian Ayala

Use the ideal gas law. Initial pressure (Pi) is 4 atm, final pressure (Pf) is 5 atm. Pressure change is 1 atm.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.