Molar Specific Heat of Ideal Gases

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1 Molar Specific Heat of Ideal Gases Since Q depends on process, C dq/dt also depends on process. Define a) molar specific heat at constant volume: C V (1/n) dq/dt for constant V process. b) molar specific heat at constant at constant pressure: C P = (1/n) dq/dt for constant P process. Consider constant V process: W = 0 and Q = E int. Q = n C V dt (= nc V T if C V = constant) Therefore E int = n C V dt Since E int /n only depends on temperature end points ( Joule effect ), E int = n C V dt for any process. Consider constant P process:. Q = n C P dt and W = - PdV. Since P = constant, dv = nr dt/p, dw = - P (nr dt/p) = -nr dt Q = E int W n C P dt = n C V dt + nr dt C P = C V + R (for any ideal gas)

2 Ideal Gas: C P = C V + R a) C P > C V : true for everything (not just ideal gas) (For liquids and solids, the difference is usually small.) b) If C V = constant, then C P = constant. i) Constant V process, Q = n C V dt = nc V T ii) Constant P process: Q = n C P dt = nc P T iii) Any process: E int = = n C V dt = nc V T c) monoatomic ideal gas: no internal excitation of molecule so E int = N<K trans > = 3/2 Nk B T = 3/2 nrt C V = 3/2 R = 12.5 J/mol K, C P = 5/2 R = 20.8 J/mol K, C P /C V = 5/3 = 1.67 a Values at T = 300 K and P = 1 atm. Small deviations of C V due to non-ideal properties interactions between molecules

3 Equipartition of Energy Theorem: The average (i.e. thermal) value of the energy for each classical, quadratic degree of freedom = ½ k B T. Consider a diatomic molecule (e.g. H 2, N 2, O 2, HCl, [since air is 98% nitrogen and oxygen, it is mostly diatomic]). In addition to CM translation (3 quadratic degrees of freedom), it has rotational and vibrational energy. K rot = ½ Iω z 2 + ½ Iω x 2 [no energy for rotation about inter-atomic (y) axis since I y = 0]: 2 quadratic degrees of freedom. E vib = K vib + U vib = ½ µ (ds/dt) 2 + ½ k (s-s 0 ) 2 effective (reduced) mass effective spring constant s Therefore, total of 7 quadratic degrees of freedom, expect E int = 7/2 Nk B T = 7/2 nrt C V = 7/2 R = 29.1 J/mol K C P = 9/2 R = 37.4 J/mol K

4 Molar Specific Heat of Hydrogen C V steps from 3/2 R (T < ~ 100 K) to 5/2 R (T < 1000 K) to 7/2 R. only translations translations + rotations translations, rotations, and vibrations

5 Hydrogen Equipartition of Energy Theorem: The average (i.e. thermal) value of the energy for each classical, quadratic degree of freedom = ½ k B T. classical: k B T >> ε = the quantum of energy Vibrations: ε vib h(k/µ) 1/2 Rotations: ε rot h 2 /I, where Planck s constant h = 1.05 x J s. For k B T < ε vib, vibrations freeze out cannot be thermally excited, and C V 5/2 R. For k B T < ε rot, rotations freeze out cannot be thermally excited, and C V 3/2 R. For nitrogen and oxygen, which are much heavier than hydrogen, rotations theoretically freeze out a few Kelvin, i.e. below boiling point, so not observable in gas, but vibrations freeze out near 1000 K (like H 2 ). Near room temperature, most diatomic molecules are on the C V ~ 5/2 R shelf.

6 Expect C P - C V = R = J/mol K 3/2 R = 12.5 J/mol K 5/2 R = 20.8 J/mol K 5/2 R = 20.8 J/mol K 7/2 R = 29.1 J/mol K F [Polyatomic may have 3 rotational and several vibrational degrees of freedom, some of which are frozen out at room temperature and some not.] [Deviations from expected values due to interactions between molecules (i.e. non-ideal behavior) and because C V = 5/2 shelf not perfectly flat.]

7 Problem: 3000 J of energy are transferred by heat into 10 moles of an ideal diatomic gas near room temperature. Find the work and changes in internal energy and temperature for the following cases: a) Constant V process b) Constant P process (isobar) c) Constant T process (isotherm)

8 Problem: 3000 J of energy are transferred by heat into 10 moles of an ideal diatomic gas near room temperature. Find the work and changes in internal energy and temperature for the following cases: a) Constant V process b) Constant P process (isobar) c) Constant T process (isotherm) Diatomic near room temp.: C V = 5/2 R, C P = 7/2 R c) T = 0 E int = 0 W = - Q = J a) V = 0 W = 0 E int = Q = 3000 J E int = n C V T = 5/2 nr T T = (2/5) (3000 J) / [(10 moles)(8.314 J/mol K)] = 14.4 K

9 Problem: 3000 J of energy are transferred by heat into 10 moles of an ideal diatomic gas near room temperature. Find the work and changes in internal energy and temperature for the following cases: a) Constant V process b) Constant P process (isobar) c) Constant T process (isotherm) Diatomic near room temp.: C V = 5/2 R, C P = 7/2 R c) T = 0 E int = 0 W = - Q = J a) V = 0 W = 0 E int = Q = 3000 J But E int = n C V T = 5/2 nr T T = (2/5) (3000 J) / [(10 moles)(8.314 J/mol K)] = 14.4 K b) Q = n C P T = 7/2 nr T T = (2/7) (3000 J) / [[(10 moles)(8.314 J/mol K)] = 10.3 K E int = nc V T = (5/2) nr T = 2143 J W = E int Q = ( ) J = J

10 γ C P / C V = (C V + R)/C V = 1 + R/C V, (γ = constant (> 1) if C V = constant). C V = 3/2 R γ = 5/3 = 1.67 (its maximum value for ideal gas in 3D) C V = 5/2 R γ = 7/5 = 1.40 C V = 7/2 R γ = 9/7 = 1.29

11 Significance of γ C P /C V Consider an adiabatic (Q = 0: well insulated and/or too rapid for heat flow, but still quasistatic ) process in an ideal gas with constant C V (e.g. on a shelf ). de int = dw nc V dt = -P dv dt = - (1/nC V ) PdV But PV = nrt nr dt = P dv + V dp (P dv + V dp)/nr = - (1 / nc V ) P dv (P dv + V dp) = - (R / C V ) P dv (1 + R/C V ) P dv = - V dp γ P dv = - V dp γ dv / V = - dp / P γ ln V = const ln P exp(γ ln V) = exp (const ln P) V γ = const/p PV γ = constant = P i V i γ (for an adiabatic process in an ideal gas with constant C V.)

12 P / P i Isotherm: PV = constant V / V i = 9/7 = 7/5 = 5/3 adiabats PV γ = constant since γ > 1, adiabats are steeper than isotherms

13 Problem: Consider a diatomic ideal gas at high temperature, so C V = 7/2 R and γ = 9/7. It is adiabatically compressed so that P f = 3 P i. What are the ratios of V f /V i and T f /T i? P f V f γ = P i V i γ V f /V i = (P i /P f ) 1/γ = (1/3) 7/9 = T f = P f V f /nr = (3P i ) (0.426 V i ) / nr = 1.28 P i V i /nr = 1.28 T i T f /T i = 1.28

14 Problem: Consider the quasistatic cycle on the right, for n moles of an ideal gas with molar specific heat ratio γ. AB is an adiabatic process. Find the work, heat, and E int for each step. [Give all results in terms of n, γ, P 0, V 0.] Preliminaries 1) Find C P and C V in terms of γ: C P = C V + R Therefore γ = (C V +R)/C V = 1 +R/C V C V = R/(γ-1) C P = γr/(γ-1) 2) Can find T s in terms of PV s, so first must find P B : P A V A γ = P B V B γ P B = P A (V A /V B ) γ = P 0 /3 γ P / P C A V / V 0 T A = P 0 V 0 /nr T B = (P 0 /3 γ )(3V 0 )/nr = P 0 V 0 /(3 γ-1 nr) T C = (P 0 /3 γ )(V 0 )/nr = P 0 V 0 /(3 γ nr) B

15 Problem: Consider the quasistatic cycle on the right, for n moles of an ideal gas with molar specific heat ratio γ. AB is an adiabatic process. Find the work, heat, and E int for each step. [Give all results in terms of n, γ, P 0, V 0.] C V = R/(γ-1) C P = γr/(γ-1) P B = P 0 /3 γ T A = P 0 V 0 /nr, T B = P 0 V 0 /(3 γ-1 nr) T C = P 0 V 0 /(3 γ nr) P / P A C B V / V 0 C A is constant volume, so W CA = 0 and Q CA = nc V (T A -T C ) = [P 0 V 0 /(γ-1)](1-1/3 γ ) (heat in) E CA = W CA + Q CA = P 0 V 0 /(γ-1)](1-1/3 γ ) B C is constant pressure, so W BC = -P B (V C -V B ) = (2P 0 V 0 /3 γ ) Q BC = nc P (T C -T B ) = - [γp 0 V 0 /(γ-1))](1/3 γ-1 1/3 γ ) = -2γP 0 V 0 /[3 γ (γ-1)] E BC = W BC + Q BC = (2P 0 V 0 /3 γ ) [1-γ/(γ-1)] = -2P 0 V 0 /[3 γ (γ-1)] (heat out) A B is adiabatic, so Q AB =0. W AB = - PdV = - P 0 (V 0 /V) γ dv W AB = -P 0 V 0 γ dv/v γ = [P 0 V 0γ /(γ-1)][1/(3v 0 ) γ-1 1/V 0 γ-1 ] = -[P 0 V 0 /(γ-1)] [1-1/3 γ-1 ] E AB = W AB + Q AB = -[P 0 V 0 /(γ-1)] [1-1/3 γ-1 ]

16 A 0.8 P / P C B V / V 0 W CA = 0, Q CA = (P 0 V 0 /(γ-1)](1-1/3 γ ), E CA = [P 0 V 0 /(γ-1)](1-1/3 γ ) W BC = (2P 0 V 0 /3 γ ), Q BC = -2γP 0 V 0 /[3 γ (γ-1)], E BC = -2P 0 V 0 /[3 γ (γ-1)] W AB = -[P 0 V 0 /(γ-1)] [1-1/3 γ-1 ], Q AB = 0, E AB = -[P 0 V 0 /(γ-1)] [1-1/3 γ-1 ] Note that E cycle = E AB + E BC + E CA = [P 0 V 0 /(γ-1)] [1-1/3 γ -2/3 γ /3 γ-1 ] = [P 0 V 0 /(γ-1)] [-3/3 γ + 1/3 γ- 1] = 0, as expected!

17 A W CA = 0, Q CA = (P 0 V 0 /(γ-1)](1-1/3 γ ), E CA = [P 0 V 0 /(γ-1)](1-1/3 γ ) W BC = (2P 0 V 0 /3 γ ), Q BC = -2γP 0 V 0 /[3 γ (γ-1)], E BC = -2P 0 V 0 /[3 γ (γ-1)] P / P C W AB = -[P 0 V 0 /(γ-1)] [1-1/3 γ-1 ], Q AB = 0, E AB = -[P 0 V 0 /(γ-1)] [1-1/3 γ-1 ] B V / V 0 e.g. if diatomic near room temperature, γ = 7/5: W CA = 0, Q CA = E CA = 1.96 P 0 V 0 W BC = 0.43 P 0 V 0, Q BC = P 0 V 0, E BC = P 0 V 0 W AB = P 0 V 0, Q AB = 0, E AB = P 0 V 0