Practice Problems from Chapter 1-3 Problem 1 One Mole of a Monatomic Ideal Gas Goes Through a Quasistatic Three-Stage Cycle (1-2, 2-3, 3-1) Shown in V 3 the Figure

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Practice Problems from Chapter 1-3 Problem 1 One mole of a monatomic ideal gas goes through a quasistatic three-stage cycle (1-2, 2-3, 3-1) shown in V 3 the Figure. T1 and T2 are given. V 2 2 (a) (10) Calculate the work done by the gas. Is it positive or negative? V 1 1 (b) (20) Using two methods (Sackur-Tetrode eq. and dQ/T), calculate the entropy change for each stage and ∆ for the whole cycle, Stotal. Did you get the expected ∆ result for Stotal? Explain. T1 T2 T (c) (5) What is the heat capacity (in units R) for each stage? Problem 1 (cont.)

∝ → (a) 1 – 2 V T P = const (isobaric process) δW 12=P ( V 2−V 1 )=R (T 2−T 1)>0

V = const (isochoric process) 2 – 3 δW 23 =0

V 1 V 1 dV V 1 T1 3 – 1 T = const (isothermal process) δW 31=∫ PdV =R T1 ∫ =R T 1 ln =R T1 ln ¿ 0 V V V 2 T 2 V 2 2

T1 T 2 T 2 δW total=δW 12+δW 31=R (T 2−T 1)+R T 1 ln =R T 1 −1−ln >0 T 2 [ T 1 T 1 ] Problem 1 (cont.)

Sackur-Tetrode equation: V (b) 3 V 3 U S ( U ,V ,N )=R ln + R ln +kB ln f ( N ,m ) V2 2 N 2 N

V f 3 T f V f 3 T f ΔS =R ln + R ln =R ln + ln V V 2 T V 2 T 1 1 i i ( i i )

1 – 2 V ∝ T → P = const (isobaric process)

T1 T2 T 5 T 2 ΔS12 = R ln 2 T 1 T T V = const (isochoric process) 3 1 3 2 2 – 3 ΔS 23 = R ln =− R ln 2 T 2 2 T 1

V 1 T 2 3 – 1 T = const (isothermal process) ΔS 31 =R ln =−R ln V 2 T 1

5 T 2 T 2 3 T 2 as it should be for a quasistatic cyclic process ΔS = R ln −R ln − R ln =0 cycle 2 T T 2 T (quasistatic – reversible), 1 1 1 because S is a state function. Problem 1 (cont.)

δ Q V (b) d S = - for quasi-static processes 3 T V2 2 1 – 2 V ∝ T → P = const (isobaric process)

T2 V1 C P dT 5 T 2 1 δ Q=C dT ΔS = = R ln P 12 ∫ T 2 T T 1 1

T1 T2 T 2 – 3 V = const (isochoric process)

T1 CV dT 3 T 2 d U =δ Q δ Q=C dT ΔS = =− R ln V 23 ∫ T 2 T T 2 1 3 – 1 T = const (isothermal process)

V 1 V 1 1 RT 1 dV V 1 T 2 d U =0 δQ=−δWON ΔS31 = ∫ PdV = ∫ = R ln =−R ln T V T V V 2 T 1 1 2 1 V 2

5 T 2 T 2 3 T 2 ΔS cycle = R ln −R ln − R ln =0 2 T1 T 1 2 T 1 Problem 1 (cont) = Let’s express both δQ and dT in terms of dV : V (c) δQ C d T 3 V2 2 1 – 2 V ∝ T → P = const (isobaric process) V 3 5 1 1 C=C =C +R= R+R= R P V 2 2

T1 T2 T 2 – 3 V = const (isochoric process) 3 C=C = R V 2 3 – 1 T = const (isothermal process), dT = 0 while δQ ≠ 0 C=∞

At home: recall how these results would be modified for diatomic and polyatomic gases. Problem 2 One mole of a monatomic ideal gas goes through a quasistatic three-stage cycle (1-2, 2-3, 3-1) shown in the Figure. Process 3-1 is adiabatic; P , V , and V are given. P 1 1 2 1 2 (a) (10) For each stage and for the whole cycle, express the δ P1 work W done on the gas in terms of P1, V1, and V2. Comment on the sign of δW. 3 (b) (5) What is the heat capacity (in units R) for each stage? (c) (15) Calculate δQ transferred to the gas in the cycle; the same for the reverse cycle; what would be the result if δQ V1 V2 V were an exact differential? (d) (15) Using the Sackur-Tetrode equation, calculate the entropy change for each stage and for the whole cycle, ∆ ∆ Stotal. Did you get the expected result for Stotal? Explain. Problem 2 (cont.) P 1 2 P1

V V 1 2 V (a) 1 – 2 P = const (isobaric process) δ W 12=−P1 (V 2−V 1 )<0

2 – 3 V = const (isochoric process) δ W 23=0 3 – 1 adiabatic process

V 1 V 1 P1 V 1γ 1 1 1 δ W 31=−∫ P (V )dV =−∫ γ dV =−P 1 V 1γ − − γ−1 γ−1 V V V 1 γ [V 1 V 2 ] 2 2 2/3 5 3 V 1 = γ= =P1 V 1 1− >0 ( 3 ) 2 ( [V 2 ] ) P 1 2 Problem 2 (cont.) P1

5 P2 V 2 P1 V 1 5 δ Q =C T −T = R − = P V −V >0 V V 12 P ( 2 1) 2 ( R R ) 2 1 ( 2 1) 1 2 V 2 – 3 V = const (isochoric process) 5 3 γ γ 3 V 1 3 δ Q23=CV (T 3−T 2 )= V 2 ( P3−P1 )=[ P1 V 1=P3 V 2]= P1 V 2 −1 <0 2 2 [(V 2 ) ] 3 – 1 adiabatic process δ Q31=0

5 5 3 V 1 3 δ Q=δ Q12+δ Q 23= P1 (V 2−V 1 )+ P1 V 2 −1 2 2 (( V 2 ) ) For the reverse cycle: δ Qreverse=−δ Q

If δQ were an exact differential, for a cycle δQ should be zero. Problem 2 (cont.)

Sackur-Tetrode equation: P (d) 1 2 3 S ( U ,V ,N )=R ln V + R ln U +k B ln f ( N ) P1 2

V f 3 T f V f 3 T f ΔS =R ln + R ln =R ln + ln 3 V 2 T ( V 2 T ) i i i i

V V ∝ → 1 2 V 1 – 2 V T P = const (isobaric process) 5 V 2 ΔS12 = R ln 2 V 1 2 – 3 V = const (isochoric process) γ 3 T 3 T 3 P3 V 1 3 V 1 5 V 1 ΔS 23 = R ln = = P3=P1 = R γ ln = R ln 2 T 2 [ T 2 P1 ( V 2 ) ] 2 V 2 2 V 2

δQ = 0 (quasistatic adiabatic = isentropic process) 3 – 1 ΔS 31 =0

5 V 2 5 V 2 as it should be for a quasistatic cyclic process ΔS = R ln − R ln =0 cycle 2 V 2 V (quasistatic – reversible), 1 1 because S is a state function. Problem 3 Calculate the heat capacity of one mole of an ideal monatomic gas C(V) in the quasi-static process shown

in the Figure. P0 and V0 are given.

P 0 δ Q 1 Start with the definition: C≡ P d T 0 3 20 δQ=dU−δW = RdT +P (V ) dV we need to find 2 the equation of δ Q 3 dV this process C (V )= = R+P(V ) d T 2 dT V=V(T)

P0 V 30 P= P − V = P 1− 0 V 0 V V 0 ( 0 ) P V 1− =RT 0 0 V V ( V 0 ) 0 PV =RT dV R 1 = dT P0 V P0 dV V P0 V dV P0 dV V 1−2 40 1− + − = 1−2 =dT ( V 0 ) R ( V ) R ( V ) R ( V ) 0 0 0 5 V R −4 3 dV 3 V V ( 2 V 0 ) C (V )= R+P(V ) = R+R 1− / 1−2 C (V )= V 2 dT 2 ( V 0 ) ( V 0 ) 1−2 V 0 P Problem 3 (cont.) P S=const 0 adiabat 50 Does it make sense?

5 V R −4 C/R ( 2 V 0 ) C (V )= V 1−2 V 0 T=const isotherm C (V )=∞ at V =V 0 /2 the line touches an isotherm

2.5 0 1.5 V0/2 5V0/8 V0 V

0 1/2 5/8 1 V/ V0 C (V )=0 at V =5V0 /8 the line touches an adiabat Problem 4 (10) The ESR (electron spin resonance) set-up can detect the minimum difference in the number of “spin-up” and “spin-down” electrons in a two-state paramagnet 10 N↑-N↓ =10 . The paramagnetic sample is placed at 300K in an external magnetic field B = 1T. The component of the electron’s magnetic moment ± µ ± -24 along B is B = 9.3x10 J/T. Find the minimum total number of electrons in the sample that is required to make this detection possible.

N E2−E1 2μ B ↓ =exp − E − E =2μ B N −N =N 1−exp − B =1010 N k T 2 1 B ↑ ↓ ↑ ↑ ( B ) [ ( kB T )]

10 2μB B 10 2μB B N ↑+N ↓=N ↑ 1+exp − = 1+exp − [ ( kB T )] 2μ B [ ( k B T )] 1−exp − B k T [ ( B )]

−24 2μ B B 2×9.3⋅10 ×1 exp − =exp − =0 .9955 - the high-T limit ( k T ) ( 1.38⋅10−23 ×300 ) B

2⋅1010 N +N ≈ =4,4⋅1012 ↑ ↓ 0.0045 Problem 5

Consider a system whose multiplicity N f N /2 is described by the equation: Ω ( U ,V , N ) =f ( N ) V U where U is the internal energy, V is the volume, N is the number of particles in the system, Nf is the total number of degrees of freedom, f(N) is some function of N. (a) (10) Find the system’s entropy and temperature as functions of U. Are these results in agreement with the equipartition theorem? Does the expression for the entropy makes sense when T → 0? (b) (5) Find the heat capacity of the system at fixed volume. (c) (15) Assume that the system is divided into two sub-systems, A and B; sub-system

A holds energy UA and volume VA, while the sub-system B holds UB=U-UA and

VB=V-VA. Show that for an equilibrium macropartition, the energy per molecule is the same for both sub-systems.

N f 1 ∂ S N f (a) S=k B lnΩ=k B ln f ( N ) +Nk B lnV + k B ln U = =k 2 T ( ∂U ) B 2U 2U N f N ,V T = - in agreement with the equipartition theorem U = k B T N f k B 2 When T → 0, U → 0, and S → - ∞ - doesn’t make sense. This means that the expression for Ω holds in the “classical” limit of high temperatures, it should be modified at low T. Problem 5 (cont.)

(b) δQ ∂ U ∂ S f 1 f CV ≡ = = [ dU=TdS−PdV ]=T =T NkB = NkB ( ∂T )N ,V ( ∂T )N ,V ( ∂ T )N ,V 2 T 2

N 3N / 2 A N B A 3NB / 2 Ω ( U ,V , N ) ∝ V A ( V −V A) U A (U −U A)

3N 3N /2−1 3N 3N / 2 3N /2−1 ∂ Ω A ( A ) 3NB / 2 A A ( B ) ∝ U A (U −U A) − U A (U −U A) =0 ∂U A 2 2

3N A 3N B U A U B (U −U A ) − U A =0 = 2 2 N A N B