Question

Calculate the change in the entropy of the system and also the change in the entropy of the surroundings, and the resulting total change in entropy, when a sample of nitrogen gas of mass 14 g at 298 K and 1.00 bar doubles its volume in (a) an isothermal reversible expansion, (b) an isothermal irreversible expansion against Pext = 0, and (c) an adiabatic reversible expansion.

Answer 1

(a) Δ
`S_(sys)` = 2.881 J/K; Δ
`S_(sur)` = -2.881 J/K; total change in entropy = 0

(b)Δ
`S_(sys)` = 2.881 J/K; Δ
`S_(sur)` = 0 ; total change in entropy = 2.881 J/K

(c) Δ
`S_(sys)` = 0 ; Δ
`S_(sur)` = 0 ; total change in entropy = 0

Step-by-step explanation:

In the given problem, we need to calculate the change in the entropy of the system and also the change in the entropy of the surroundings, and the resulting total change in entropy, when a sample of nitrogen gas of mass 14 g at 298 K and 1.00 bar doubles its volume. We have the following variable:

mass (m) = 14 g

Temperature = 298 K

Pressure = 1.00 bar

Initial volume =
`V_(1)`

Final volume =
`V_(2)` =
`2V_(1)`

(a) Change in entropy of the system Δ
`S_(sys)` =
`nRIn(V_(2) )/(V_(1) )`

where R = 8.314 J/(mol*K)

n = number of moles = mass/molar mass = 14/ 28 = 0.5 moles

Δ
`S_(sys)` = 0.5*8.314*ln2 = 2.881 J/K

Change in entropy of the surrounding Δ
`S_(sur)` = -2.881 J/K

Therefore, for a reversible process, the total change in entropy = Δ
`S_(sys)`+Δ
`S_(sur)` = 2.881 - 2.881 = 0

(b) Because entropy is a state function, we use the same procedure as in part (a). Thus, Δ
`S_(sys)` = 2.881 J/K

Since surrounding does not change in this process Δ
`S_(sur)` = 0.

total change in entropy = Δ
`S_(sys)`+Δ
`S_(sur)` = 2.881 - 0 = 2.88 J/K

(c) For an adiabatic reversible expansion, q(rev) = 0, thus:

Δ
`S_(sys)` = 0

Since heat energy is not transferred from the system to the surrounding

Δ
`S_(sur)` = 0

total change in entropy = Δ
`S_(sys)`+Δ
`S_(sur)` = 0