Can The Entropy Of An Ideal Gas Change During An Isothermal Process?

Entropy involving ideal gases

Trouble:

Calculate the entropy change of an ideal gas that undergoes a reversible isothermal expansion from volume V_one to Five_ii.

Solution:

Concepts:
Isothermal processes
Reasoning:
For an ideal gas PV = nRT.
For an isothermal process PV = constant, dU = dQ - dW = 0. dQ = dW = PdV.
Details of the adding:
dS = dQ/T = PdV/T. ΔS = (1/T) ∫₁ ²PdV = (nR) ∫₁ ^two(one/5)dV = nRln(V_ii/V_i).

Problem:

Calculate the entropy modify of 1 mole of an ideal gas that undergoes an isothermal transformation from an initial land of pressure one.5 atm and a book of 500 cm^three to a concluding state of pressure 0.90 atm.

Solution:

Concepts:
Isothermal processes
Reasoning:
For an ideal gas PV = nRT.
For an isothermal process PV = constant, dU = dQ - dW = 0, dQ = dW = PdV.
Details of the calculation:
dS = dQ/T = PdV/T. ΔS = (1/T) ∫_A ^BPdV = (nR) ∫_A ^B(i/V)dV = nRln(V_B/V_A)
= nRln(P_A/P_B).
Hither north = 1, ΔS = viii.31 * ln(1.5/0.9) J/Chiliad = 4.ii J/K.

Problem:

Ane mole of an ideal gas is compressed at sixty ^oC isothermally from 5 atm to xx atm.
(a) Find the piece of work done.
(b) Observe the entropy change for the gas and interpret its algebraic sign
Gas abiding : 8.314 J/(mol Grand)

Solution:

Concepts:
Ideal gas law: PV = nRT, work washed on the system: Westward = -∫PdV
Energy conservation: ΔU = ΔQ + ΔW
Change in entropy: ΔS = ∫_i ^f dS = ∫_i ^f dQ_r/T
The subscript r denotes a reversible path.
Reasoning:
Using the ideal gas constabulary we tin observe the work done on the arrangement. Using ΔU = ΔQ + ΔW we tin can then find the amount of oestrus transferred to or from the arrangement. Knowing ΔQ allows united states of america to calculate ΔS.
Details of the calculation:
(a) The temperature is abiding:
-∫PdV = -nRT∫(1/5)dV = nRT ln(V_i/5_f) = nRT ln(P_f/P_i) = RT ln(4)
West = (8.314 J/K) * (333 K) ln(4) = 3838 J
(b) Free energy conservation:
increase in internal free energy of a system
= heat put into the organization + work done on the arrangement by its surroundings,
or ΔU = ΔQ + ΔW.
The temperature stays constant, the internal energy stays abiding. The work washed on the system is positive, therefore ΔQ = - ΔW is negative.
ΔS = ΔQ/T = (-3838 J)/(333 One thousand) = -xi.5 J/One thousand

Problem:

The figure below shows a maximally efficient Carnot wheel in the pressure-volume plane for a estrus engine operating between ii estrus reservoirs to perform work.
(a) For each label 1 through 4 identify the process, say whether work is done by the working fluid or on it and whether rut is added to the working fluid or extracted from it.
(b) Make a diagram showing the same bicycle in the temperature-entropy plane. Characterization the parts of your diagram that correspond to the parts labeled in the P-V diagram and put arrows on each segment indicating the direction it is traversed.

Solution:

Concepts:
The Carnot cycle, isothermal and adiabatic processes, entropy.
Reasoning:
We are asked to depict the Carnot cycle in diverse ways.
Details of the calculation:
(a)
(i) Isothermal expansion: The engine is in contact with a rut reservoir at temperature T_ane.
The engine does work. ΔW = ∫PdV. Heat is added to the working fluid.
(2) Adiabatic expansion: During the expansion the temperature falls to T_ii.
The engine does work. ΔW = ∫PdV. No oestrus is added or extracted from the working fluid.
(three) Isothermal compression: The engine is in contact with a rut reservoir at temperature T₂.
Work is washed on the engine. ΔW = -∫PdV. Heat is extracted from the working fluid.
(4) Adiabatic compression: During the pinch the temperature rises to T_i.
Work is done on the engine. ΔW = -∫PdV. No heat is added or extracted from the working fluid.
(b)
Modify in entropy: ΔS = ∫_i ^f dS = ∫_i ^f dQ_r/T
The subscript r denotes a reversible path.

Problem:

A cylinder is partitioned by a membrane into a volume V₁ initially filled with a classical platonic gas of N particles with no internal degrees of freedom at temperature T, and a book V₂ initially enclosing a perfect vacuum.
(a) The cylinder is in contact with a rut reservoir at temperature T. The membrane is moved slowly without friction, assuasive the gas to fill the entire cylinder. Compute the work done by the gas, the heat transferred betwixt the gas and the heat bath, and the change in the entropy of the gas. Is this a reversible process?
(b) The cylinder is returned to its initial state and insulated from the estrus bath. The membrane is allowed to break, releasing the gas to fill the entire book. Assume that the expansion occurs essentially instantaneously, and a new equilibrium is reached. Compute the work done by the gas and the alter in the entropy. Is this a reversible procedure?

Solution:

Concepts:
Entropy, energy conservation, the platonic gas law
Reasoning:
Modify in entropy: ΔS = ∫_i ^f dS = ∫_i ^f dQ_r/T, where the subscript r denotes a reversible path.
Isothermal expansion can be a reversible process. For isothermal expansion ΔS = ΔQ_r/T.
Nosotros detect ΔQ using energy conservation and the platonic gas constabulary.
Details of the adding:
(a) The work washed by the gas is West = ∫PdV.
PV = NkT. The temperature is constant so P = constant/5.
W = NkT∫(1/V)dV = NkT ln(Five_f/5_i) = NkT ln(1 + V₂/V₁).
ΔU = ΔQ - ΔW. U = internal energy, Q = heat put into the system, Due west = work done by the system.
ΔU = 0, since the temperature is constant.
ΔQ = ΔW = NkT ln(1 + 5₂/V_ane).
ΔS = ΔQ_r/T = Nk ln(1 + V₂/V_ane). This is a reversible process.
(b) The temperature is proportional to the average kinetic energy of the atoms. The average kinetic energy of the atoms does non change, then T stays constant. No piece of work is done by the gas, no heat is transferred to or from the gas. The change in entropy depends only on the initial and final state. These states are the same as in function (a).
ΔS = Nk ln(1 + V₂/Five₁).
This is not a reversible process.

Problem:

Consider a organisation of one mole of an ideal gas A and 3 moles of an ideal gas B at the same pressure level P and temperature T, in volumes of 5_A and V_B respectively. The two gases are separated past a partition so they are each sequestered in their corresponding volumes. If the sectionalisation is removed, calculate the change in entropy of the arrangement.

Solution:

Concepts:
Entropy, the platonic gas police
Reasoning:
Change in entropy: ΔS = ∫_i ^f dS = ∫_i ^f dQ_r/T, where the subscript r denotes a reversible path.
The gases will mix. To summate the entropy change, we treat the mixing as two separate gas expansions, 1 for gas A and another for gas B.
Here nosotros have and expansion at constant temperature. For a reversible path leading from the initial to the terminal state for each gas
dU = 0, dQ = PdV = nRTdV/V, dS = nRdV/V.
Details of the calculation:
For each gas ∆S = nRln(V_f/5_i). Then for the two dissever gas expansions we take
∆S_A = n_ARln((V_A + V_B)/V_A), ∆Due south_B = n_BRln((Five_A + 5_B)/V_B).
The full entropy change is
∆Southward = ∆South_A + ∆S_B = north_ARln((V_A + V_B)/V_A) + n_BRln((V_A + V_B)/V_B).
PV = nRT implies that the initial volume is directly proportional to the number of moles. Therefore ∆Due south = n_ARln((n_A + n_B)/northward_A) + n_BRln((n_A + n_B)/n_B).
∆S_mix = Rln(4) + 3Rln(iv/3) = 2.25 R = 18.7 J/1000.