Tuesday, 11 May 2010

Important Thermal Processes

ISOBARIC Pressure is Constant (ΔP = 0)

 Example: Gas Heated in a Cylinder fitted with a movable frictionless piston. The pressure the atmosphere and the pressure due to the weight of the piston remains constant as the gas heats up and expands. First Law Implications: ΔU = Q - W Unlike some of the other processes below neither the heat Q , the work W, or the change in internal energy ΔU are necessarily zero in a constant pressure process. For an ideal gas, constant pressure work is easily determined, W = ∫ PdV = PΔV Part of the heat that flows into the system causes the temperature to rise, Q = n cp ΔT = m Cp ΔT, the rest goes into work.
See Summery Table for more change for an Ideal gas.
ISOTHERMAL Temperature is Constant (ΔT = 0)

 Example1: Boiling of water in the open air. In general most isobaric phase changes are isothermal. In this example the system does work as the steam-produced pushes against the atmosphere as it expands. Neither the heat Q , the work W, or the change in internal energy ΔU are zero. In this case Q = mLv since the water changes phase. Example 2: In general for an Ideal gas U is only a function of the temperature so that ΔU is always equal to zero for an isothermal process.Since ΔU = 0 then W = Q from the First Law. What has to happen for this process to be isothermal is that the gas in a cylinder is compressed slowly enough that heat flows out of the gas at the same rate at which is being done on the gas. The ideal gas law can be used to determine the work done W = PV ln(Vf/Vo) which is also the equation for Q. Note that P1V1 = P2V2 = nRT, the ideal gas law for an isothernal process.

ISOCHORIC (Isovolumetric) Volume is constant (ΔV = 0)

 Example: Heating of a Gas in a Rigid, Closed container. In this case no work is done on the gas because W = ∫ PdV = ∫ P 0 = 0. As a result the FirstLaw implies that the change in internal energy must equal any heat flowing into or out of the system, ΔU = Q = n cv ΔT = m Cv ΔT. Note that V1 = V2 = nRT1/P1 = nRT2/P2, the ideal gas law for constant volume process.

ADIABATIC No heat flows into or out of the system (Q = 0)

 Example Compression of a Gas in an Insulated Cylinder. In this case any change in the internal energy of the gas is due to work done on it or by it, ΔU = W. Normally if ΔU changes the temperature of a system will change. Any temperature rise or fall is due to the work done or by the gas alone and not due to heat flowing into or out of the system since Q = 0. If a process is carried out fast enough the heat flow will be small and the process can be approximate as being adiabatic. This happen because heat flow is in general a slow process. Observe that we did not say that Q is constant because it not a state variable. Q represent an energy transfer not the heat energy of the system. In addition to the ideal gas law PV = NkT, the quantity PVγ is constant for an ideal gas where γ = cP/cV, the ratio of molar specific heats. For an ideal gas the work W = (P1V1 - P2V2)/(γ - 1)

ISOENTROPIC Entropy is constant (ΔS = 0)

 Example: A Heat Engine in which the working fluid undergoes an Adiabatic Reversible cyclic process. Any isoentropic process is also adiabatic since ΔU = ∫ dQ/T and Q = 0. However, not all adiabatic process are isoentropic. Adiabatic free expansion is not isoentropic. For a heat engine to be reversible, not only must the change in entropy of the working fluid be zero but also the net change of the entropy of the environment (heat reservoirs) must also be zero.

POLYTROPIC PVn is constant

 Here n can be any real number. All the process above can thought of as a special case of a polytropic process for an ideal gas. n = 0 for Isobaric process since PV0= P = constant. n = 1 for Isothermal process since PV1 = PV = NKT = constant. n = for Isovolumetric process (this one is not so obvious) n = γ for Adiabatic process (again not so obvious) Example: Compression or Expansion of a Gas in a Real System such as a Turbine. from from