# Calculate ΔS (for the system)

Calculate Δ*S* (for the system) when the state of 3.10mol of perfect gas atoms, for which*C*_{p}_{,m} = 5/2 *R*, ischanged from 25°C and 1.00 atm to 125°C and 5.00 atm.How do you rationalize the sign of Δ*S*?

a. Though Δ*S* (system) is positive, the process canstill occur spontaneously if Δ*S* (total) is negative.

b. Though Δ*S* (total) is negative, the process can stilloccur spontaneously if Δ*S* (system) is positive.

c. Though Δ*S* (system) is negative, the process can notoccur spontaneously if Δ*S* (total) is positive.

d. Though Δ*S* (system) is negative, the process canstill occur spontaneously if Δ*S* (total) is positive.

Answer:

~~~~~~~~

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 Δ*S* (system) is positive, the process canstill occur spontaneously if Δ*S* (total) is negative., thisis because there must be a increase in entropy and hence delta Shave 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 forthe reaction to occur.

c.The total entropy change is negative and so the reactioncannot occur.

d. As the temperature decreases, S_{surr} increases, therefore, the lower thetemperature, the more likely it is for the reaction to occur.