Entropy. The second law of thermodynamics Teacher guide

Theoretical material

A thermodynamic process is reversible if the process can return back in such a that both the system and the surroundings return to their original states, with no other change anywhere else in the universe. It means both system and surroundings are returned to their initial states at the end of the reverse process.

The process is internally reversible if no irreversibilities occur within the boundaries of the system. In these processes, a system undergoes through a series of equilibrium states, and when the process reverses, the system passes through exactly the same equilibrium states while returning to its initial state.

In externally reversible process no irreversibilities occur outside the system boundaries during the process. Heat transfer between a reservoir and a system is an externally reversible process if the surface of contact between the system and reservoir is at the same temperature.

A process can be reversible only when its satisfying two conditions:

·         Dissipative force must be absent.

·         The process should occur in infinite small time.

Irreversible processes are a result of straying away from the curve, therefore decreasing the amount of overall work done. An irreversible process is a thermodynamic process that departs from equilibrium. In terms of pressure and volume, it occurs when the pressure (or the volume) of a system changes dramatically and instantaneously that the volume (or the pressure) do not have the time to reach equilibrium.

A classic example of an irreversible process is allowing a certain volume of gas to release into a vacuum. By releasing pressure on a sample and allowing it to occupy a large space, the system and surroundings are not in equilibrium during the expansion process.

Here little work occurs. However, there is a requirement of significant work, with a corresponding amount of energy dissipation as heat flows to the environment. This is in order to reverse the process.

Worksheet answers and solutions

Task 1

I)                   PV = nRT

n =

n =

n ≈ 60 mole

II)                 =

P₁ = P₂ = const

V₂ = V₃ = 2V₁

T₂ = T₁·

T₂ = 800 K·

T₂ = 1600 K

III)              V₃ = 2V₁

V₃ = 10 m³

IV)             W = P·(V₂ – V₁)

W = ·(10-5) m³

W = 400 kJ

The answer to the question is NO, since 400 kJ is work done during process 1→2 and there is also work done on the system during process 3→1.

Task 2

A)    ΔS₁ =

ΔS₁ =

ΔS₁ ≈ -2.83 J/K

B)    ΔS₂ =

ΔS₁ =

ΔS₁ ≈ 3.41 J/K

C)    ΔS_t = ΔS₁ + ΔS₂

ΔS_t = 0.58 J/K

Task 3

1)      b

2)      a

3)      a

Task 4

1)      ΔS =

 = mL = nL_n

n =

M(Cu) = 64·10^-3 kg/mole

ΔS =

ΔS ≈ 9.5 J/K

Task 5

1)      False (decrease by 500 J)

2)      True

3)      True

4)      True

5)      True