The First Law of Thermodynamics 200613814 Ha Seong Hyeun 200713047 Oh Kyeong Ju

contents The First Law of Thermodynamics 2. Joule’s Law 3. Specific Heats 4. Enthalpy

Thermal energy External work Changes in temperature or change of state of the object causing the thermal form of energy. It can receive by thermal conduction and radiation. External work The work done by a system in expanding against forces exerted from ouside.

Internal energy Internal energy is the total energy contained by a thermo-dynamic system. This has two major components, kinetic energy and potential energy. The kinetic energy is due to the motion of the system's particles and the potential energy is associated with the static constituents of matter

What is the first law of thermodynamic? It is combined variably by a thermal energy & external work & internal energy. It can be written as following q – w = u₂ – u₁ differential dq – dw = du Function of state

What is the first law of thermodynamic? dW = F dx Because F = pA (A is the cross-sectional area) dW = pA dx = p dV Working substance Piston F dx

What is the first law of thermodynamic? P The work done by the substance when its volume increses by a small increment dV is equal to the pressure of the substance multiplied by its increase in volume, which is equal to the red-shaded area in the graph A P₁ P p Q B p₂ V V₁ V V₂ dV W = ∫ pdV v2 v1

What is the first law of thermodynamic? dq – dw = du dq = du + dw = pdα dw = pA dx = pdV dq = du + pdα

3.3-1 Joule’s Law Joule concluded that when a gas expands without doing external work and without taking in or giving out heat the temperature of the gas does not change (without doing external work) dw=0 (without taking in or giving out heat) dq=0 du = dq – dw therefore du=0

3.3-1 Joule’s Law The internal energy of an ideal gas is independent of its volume if the temperature is kept constant This can be the case only if the molecules of an ideal gas do not exert forces on each other In this case, the internal energy of an ideal gas will depend only on its temperature

exercise

Why rebound from the walls with lower velocities than balls struck them? Suppose that the walls of the court are permitted to move outward when subjected to a force. The force on the walls is supplied by the impact of the balls, and the work required to move the walls outward comes from a decrease in the kinetic energy of the balls. The work done by the system by pushing the walls outward is equal to the decrease in the internal energy of the system.

Specific Heats The amount of heat per unit mass required to raise the temperature by one degree Celsius. T+dT T The ratio dq/dT is called the specific heat of the material. Suppose a small quantity of heat dq is given to a unit mass of a material and ,as a consequence the temperature of the material increases from T to T+dT with out any changes in phase occurring within the material. dq dq dq The specific heat defined in this way could have any number of values, depending on how the material changes as it receives the heat.

cv = (du/dT) cv = (dq/dT)v const dq = cvdT + pdα Specific Heat at constant volume cv cv = (dq/dT)v const Volume of material is constant dq = du cv = (du/dT) * The first law of thermodynamics for an ideal gas dq = cvdT + pdα

cp = (dq/dT)p const Specific Heat at constant pressure cp The material is allowed to expand as heat is added to it and its temperature rises, but its pressure remains constant. dq = CvdT + pdα dq = CvdT + d(pα) - αdp dq = (Cv +R )dT- αdp dq = CvdT + pdα From the Ideal gas equation Pα = RT, d(pα) = RdT. At constant pressure the last term in vainishes; Cp = (dq/dT)p dq = (Cv +R )dT- αdp * Obtain an alternate form of the first law of thermodynamics; differentating Pdα + αdp = RdT Pdα = RdT – αdp Pdα = d(pα) – αdp Dq = (Cv + R)d – αdp Cp = Cv + R dq = CpdT - αdp

Enthalpy △q = (u2 - u1 ) + p(α2 – α1) = (u2 + pα2) – (u1 + pα1) The internal energies for a unit mass of the material. The work done by a unit mass of the material.

△q = h2 – h1 h ≡ u + pα dh = du + d(pα) At constant pressure, Enthalpy is defined △q = h2 – h1 h ≡ u + pα h is the enthalpy of a unit mass of the material. differentiating dh = du + d(pα)

dh = du + d(pα) dq = dh - αdp cv = du/dT du = cvdT (1) dq = cvdT + d(pα) – αdp (2) We obtain dh = du + d(pa). Substituting for du from equation (1) and combining with equation (2), we obtain it (dq = dh – adp) It equation is another form of the first law of thermodynamics. dq = dh - αdp

h corresponds to the heat required to raise dq = cpdT - αdp dq = dh - αdp dh = cpdT integrate h = cpT =0 =0 By comparing two equations, we obtain dh = cpdT In integrated form, h = cpT. If T = 0, h is taken as zero. So h corresponds to the heat required to raise the temperature of a material from 0 to T K at constant pressure. h corresponds to the heat required to raise the temperature of a material from 0 to T K at constant pressure.

dq = dh = cpdT dq = d(h + Φ) = d(cpT+ Φ) dΦ ≡ gdz = -αdp dq – dh –αdp h = cpT If the material is a parcel of air with a fixed mass that is moving about in an hydrostatic atmosphere, the quantity (h + Φ) which is called the dry static energy, is constant provided the parcel neither gains nor loses heat. (dq=0) dq = d(h + Φ) = d(cpT+ Φ) The dry static energy (dq=0)