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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

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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 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 3 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 (c) 1 – 2 P = const (isobaric process) 3 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.
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