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Phys 2101 Gabriela González

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pV=nRT

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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

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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!

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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

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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=nRT

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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!?

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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=6

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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

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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 l

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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]

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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!!

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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 W

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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

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ΔS = Sf − Si =dQTi

f

∫

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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

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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   = nCV dT/T + pdV/(pV/nR)   = nCV dT/T + nRdV/V

  ΔS = ∫dQ/T   = nCV ln (Tf/Ti) + nR ln (Vf/Vi)

Eint = (f/2)nRT = n CV T W = ∫ p dV pV = nRT

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Ideal heat engines use a cycle of reversible thermodynamic processes. A heat engine transforms energy extracted as heat from thermal reservoirs, into mechanical work. Consider a Carnot engine: a cycle with two isothermal processes at a high temperature TH (a→b) and and a low temperature TL (c →d), and two adiabatic processes (b→c, d→a).   ΔEint = 0 = Q – W → W = Q = |QH|-|QL|   ΔS = 0 = |QH|/TH - |QL|/TL

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We use the heat QH to get work W done, so efficiency is defined as   ε = |W|/|QH|  For a Carnot engine,   W = |QH|-|QL|   |QH|/TH = |QL|/TL  so εC = (|QH|-|QL|)/ |QH| = 1- |QL|/|QH| = 1- TL/TH < 1

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Three ideal Carnot engines operate between (a) 400K and 500K, (b) 500K and 600K, and (c) 400K and 600K.

1. Rank them according to their efficiencies, greatest first.

2.  If they all extract the same amount of energy per cycle from the high temperature reservoir, rank them according to the work per cycle done by the engines, greatest first.

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Three ideal Carnot engines operate between (a) 400K and 500K, (b) 500K and 600K, and (c) 400K and 600K.

1. Rank them according to their efficiencies, greatest first. ANS: ε=1-TL/TH : εc > εa > εb

a)  20% (1-4/5=0.2) b)  16% (1-5/6=0.16) c)  33% (1-4/6=0.33)

2.  If they all extract the same amount of energy per cycle from the high temperature reservoir, rank them according to the work per cycle done by the engines, greatest first. �ANS: ε = |W|/|QH| , so ranking according to W is the same as according to efficiency (above).