11/18/10 1 Phys 2101 Gabriela González 2!pV=nRT11/18/10 2 3!The average translational kinetic energy of molecules in a gas is related only to temperature (not pressure, or volume). The internal energy of a gas is the total kinetic energy, also related just to temperature. For a monoatomic, ideal gas, Eint = nNA Kavg = (3/2) N k T = (3/2) nRT 4!We defined heat capacity C and specific heat c (heat capacity per mass) as: Q = C ΔT = c m ΔT For water, c = 1cal/goC = 4187 J/kg K The molar specific heat is the heat capacity per mole. However, the amount of heat needed to raise the temperature of 1 mole by 1 kelvin depends on whether we perform the operation at constant pressure or at constant volume: Q = n CP ΔT or Q = n CV ΔT In both cases, Q= ΔEint +W: ΔEint is the same, but W is not!11/18/10 3 5!Consider a process at constant volume, raising the temperature by ΔT . Q = n CV ΔT Q= ΔEint + W W = 0 ! Q = ΔEint = (3/2) nR ΔT = n CV ΔT → CV= (3/2)R = 12.5 J/molK → Eint = n CV T Eint = (3/2)nRT 6!Consider a process at constant pressure, raising the temperature by ΔT. Q = n CP ΔT Q= ΔEint + W W = p ΔV = n R ΔT ΔEint = (3/2) nR ΔT Q = ΔEint + W = (5/2) nR ΔT → CP= (5/2)R = CV + R pV=nRT11/18/10 4 7!For any process (1, 2, 3 or 4): pV = n R T#ΔEint = Q – W Eint= (3/2) n R T = n CV T Also, CP = CV + R ; CV=(3/2)R Constant volume (4):  W=0,  Q = n CV ΔT  ΔEint= n CV ΔT Constant pressure (1):  W= p ΔV = nRΔT  Q = n CP ΔT  ΔEint= n CV ΔT Constant temperature (2):  W = nRT ln (Vf/Vi),  ΔEint= 0  Q = W = nRT ln (Vf/Vi) True for monoatomic gases, but not for others!? 8!We used Eint = nNA Kavg = nNA (3/2) kT = (3/2) n R T Is translational kinetic energy the only kinetic energy in molecules? Not for polyatomic molecules! They can spin and rotate, in a number of different ways. Every kind of molecule has a certain number f of degrees of freedom in which the molecule can store energy. Each d.o.f. has, in average, (1/2) kT kinetic energy per molecule (or (1/2)R T per mole). Thus, Eint = nNA Kavg = (f /2) nRT and CV = (f /2) R Cp = CV + R γ = Cp/CV = 1 + 2/f Cp, CV and γ can easily be measured. f=3 f=5 f=611/18/10 5 9!If volume is constant (“isochoric” process), W=0, and Tp-1 is constant. If temperature is constant (“isothermal” process), ΔEint=0, and pV is constant. If pressure is constant (“isobaric” process), W, Q, and ΔEin are not zero, but TV-1 is constant. Adiabatic processes happen when Q = 0: the system is thermally insulated, or because the process happens very quickly. Temperature, pressure and volume all change, but the quantities pVγ and T Vγ-1 are constant, where γ = CP/CV (=4/3=1.67 for monoatomic gases) The curve in a pV diagram is called an “adiabat” pV=nRT 10!One liter of gas with γ = 1.3 is at 292 K and 1.4 atm pressure. It is suddenly compressed (adiabatically) to half its original volume. Find its final pressure and temperature. Adiabatic process: pVγ and T Vγ-1 are constant: p2 = p1 (V1/V2)γ = 1.4 atm 21.3 = 3.4 atm T2 = T1 (V1/V2)γ-1 = 292K 20.3 = 359 K The gas is now cooled back to 292 K at constant pressure. What is its final volume? Constant pressure: p= nRT/V , or T/V, is constant V3=V2 (T3/T2) = 0.5 l (292 K/ 359 K) = 0.41 l11/18/10 6 11!One mole of an ideal diatomic gas undergoes a transition from a to c along the diagonal path in the figure. The temperature of the gas at point a is 1200 K. (a) What is the change in internal energy? [-5000 J] (b) How much heat is added to the gas? [2000J] (c) How much heat is added to the gas if it goes back to the original state a through the point b? [-5000J] 12!11/18/10 7 13!A free expansion process happened when Q = W = 0, for an isolated system. Since internal energy is constant, temperature is constant. Pressure and volume change, but the quantity pV = nRT is constant The curve in a pV diagram is an isotherm (but no work is done or heat is exchanged in the process!). This is a non-reversible process!! 14!We can easily tell the direction of the “arrow of time” in irreversible processes which, if spontaneous, only happen one way: • spilling fluids • gas expansion • breaking solids Conservation of energy doesn’t forbid these processes; there is another quantity, “entropy”, which measures the degree of “disorder”: S = k log W11/18/10 8 15!If an irreversible process occurs in a closed system, the entropy S of the system always increases; it never decreases. If any process occurs in a closed system, the entropy of the system increases for irreversible processes and remains constant for reversible processes. It never decreases! ΔS ≥ 0 16!ΔS = Sf− Si=dQTif∫11/18/10 9 17!An 8.0g ice cube at -10oC is put in a Thermos flask containing 100 cm3 of water at 20oC. The specific heat of ice is 2220 J/kgK, the specific heat of water is 4190 J/kgK, and the heat of fusion of ice-water is 330 kJ/kg. What’s the final equilibrium temperature? Use ΔQ = 0 = cWmW(Tf-20oC) + cice mice (0oC-(-10oC)) + mice Lf + cW mice (Tf-0oC) Answer: 12.24oC= 285.39 K What is the change in entropy of the ice-water system? Use ΔS= ∫dQ/T = ∫mcdT/T = mc ln(Tf/Ti) for changes in temperature, and ΔS= ∫dQ/T =Q/T = mLf/T for changes of phase Answer: ΔSw= -11.24 J/K ΔSice = +11.88 J/K ΔS = 0.64 J/K 18!Any isothermal process: ΔS= ∫dQ/T= (1/T) ∫dQ = ΔQ/T Any reversible process: dQ = dEint + dW = nCV dT + p dV dQ/T= nCV dT/T + pdV/T