Question

Calculate ΔS (for the system) when the state of 3.10 mol of perfect gas atoms, for which C_p,m = 5/2 R, is changed from 25°C and 1.00 atm to 125°C and 5.00 atm.


How do you rationalize the sign of ΔS?

a. Though ΔS (system) is positive, the process can still occur spontaneously if ΔS (total) is negative.

b. Though ΔS (total) is negative, the process can still occur spontaneously if ΔS (system) is positive.

c. Though ΔS (system) is negative, the process can not occur spontaneously if ΔS (total) is positive.

d. Though ΔS (system) is negative, the process can still occur spontaneously if ΔS (total) is positive.

Answer 1

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delta S = delta S1 + delta S2

For the first step

delta S1 = Cpm ln Tf/Ti

delta S1 = 3.10 * 5/2 * 8.314 JK-1 mol-1 * ln (125+273) K / (25+273)K

              = 18.7 J K-1

and for the second

delta S2 = q _rev / T

q_rev = -w = nRT ln pi/pf

delta S2= nR ln pi/pf = 3.10 * 8.314 * ln 1/5

             = -41.48 J K-1

therefore delta S = 18.7 + (-41.48)

                           delta S = - 22.78 JK-1

a. Though ΔS (system) is positive, the process can still occur spontaneously if ΔS (total) is negative., this is because there must be a increase in entropy and hence delta S have negative sign and more absolute value than delta S system .

b. As the temperature increases, S_surr increases (becomes less negative), therefore, the higher the temperature, the more likely it is for the reaction to occur.

c.The total entropy change is negative and so the reaction cannot occur.

d. As the temperature decreases, S_surr increases, therefore, the lower the temperature, the more likely it is for the reaction to occur.