Specific Heat of Diatomic Gases and Solids The Adiabatic Process Ron Reifenberger Birck Nanotechnology Center Purdue University February 22, 2012 Lecture 7 1
Specific Heat for Solids and Diatomic i Gasses In this lecture, you will learn what determines the specific heat of diatomic gasses and elemental solids. The physics behind an adiabatic gas process will also be analyzed. 2
From last lecture, you learned how to evaluate the heat capacity for an ideal monatomic gas 1 2 1 KE K mv av av av 3 kt ( for one molecule) 2 2 The internal energy of N atoms in an ideal monatomic gas is an intrinsic, distributed property of the system For ideal gas of N atoms: E int KE = 3(½ NkT) = 3(½ nrt) int C V de int dt = 3/2 nr monatomic gas equipartition theorem C P = C V + nr = 5/2 nr 3 What makes diatomic molecules different?
Rotational motion - another way to distribute energy? (rad/s) r Velocity v Mass m KE = ½ mv 2 = 2f v = r For now, no linear translation KE = ½ mr 2 2 = ½ I 2 Typical value: CO, I 1.45 x 10-46 kg m 2 4
By way of review,... Displacement Linear Motion Rotational Motion d = v o t+½ at 2 = o t + ½ t 2 Velocity v = v o + at = o + t Inertia m I Newton s 2 nd Law Momentum Conservation of momentum Kinetic Energy F = m a p = mv If F=0, then p = constant ½ mv 2 = I L = I If =0, then L = constant ½ I 2 5
For diatomic molecules, additional energy can be disbursed in rotational motion 5 degrees of ffreedom translation + rotation KE = ½mv 2 + 2 + 2 + x 2 x + y 2 x ½mv y ½mv z ½I ½I y E int = 5 (½ nrt) C v = 5/2 nr C P = 7/2 nr 6
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Any other way to distribute energy? vibrational motion Work Done = Change in Elastic Energy W = F d spring constant k (N/m) +Force = ½ kx x k x=0 x final F restore = -kx final Work done is Work done is equal to the area 8
It is possible that diatomic molecules may vibrate! Usually, this occurs at temperatures much higher than room temperature. If vibration becomes important, then additional energies must be added to heat capacity to include this possibility. 1 mole of H 2 gas 7 2 R 5 2 R 3 2 R 9
To calculate the specific heat for ELEMENTAL solids, we need a model that describes how atoms interact with each other KE = ½mv 2 x +½mv 2 y +½mv 2 z + ½kx 2 +½ky 2 +½kz 2 6 degrees of freedom E int = 6 (½ nrt) for one mole, n=1 c (molar specific heat) = 3R k = effective spring constant [N/m] c 24.9 J/mole K Law of Dulong-Petit 10
Elemental Solid C V (J/mole K) Aluminum 23.4 Bismuth 25.3 Copper 23.8 Gold 24.5 Platinum 25.4 Silver 24.4 Tungsten 24.44 Dulong-Petit value c 24.9 J/mole K 11
A Thermodynamic Process: How you go from initial iti to final state t P (P 2, V 2, T 2 ) FINAL 2 STATE 4 3 1 (P 1, V 1, T 1 ) INITIAL STATE TE V There are many ways to go from (P 1,V 1,T 1 ) to (P 2,V 2,T 2 ) 12
How to measure P-V for a gas? 13
The Adiabatic Process for an Ideal Gas is the Greek word for impassable Adiabatic Thermodynamic Process fast! insulation Quasi-static Adiabatic Process slow! no time for heat to flow in or out no heat flows in or out E int = Q in + W on 14
Using the 1 st Law for Ideal Monatomic Gas Process E int Q in W on = Isobaric nc' V T nc P T -P V Adiabatic nc' V T 0 E int Isochoric nc' V T E int 0 Isothermal 0 -W -nrt ln(v f /V i ) General nc' V T EE int -W (PV area) 15
Adiabatic Process ONLY E int = Q in + W on nc V dt= C V dt = 0 + (-PdV) C V dt = -nrt dv/v dt = - T nr dv C V V dv 1 ln( V ) C ln( V ) C ln C V V 1 T ln( T ) ln( ) ln ln TV C V 1 V NOTE: the final result follows directly from the 1 st Law ln (TV nr/c V ) = constant Key Idea: the quantity on the left is Key Idea: the quantity on the left is constant during an adibatic process involving an ideal gas! 16
Rewrite: nr C P - C V = = -1 C P /C V C V C V TV -1 = constant if T,V are known rewrite, using the ideal gas law: PV -1 = nr V constant if the number of moles is constant during the process, then PV = some other constant if P,V are known 17
Adiabatic Compression PV = constant Monatomic gas Diatomic gas γ 5/3 1.667 7/5=1.400 18
Calculating the Work in an Adiabatic Compression deint dqin dwon 0 dw on deint dq in dw de C dt adiabatic T f int W C dt C T T v only depends on ΔT! adiabatic v v f i T using PV nrt T W adiabatic C v i PV PV nr PV f f i i PV f f i i 1 nr PV nr since nr C V = -1 19
Example: If 1 mole of an ideal diatomic i gas, initially iti at 310 K, expands adiabatically from 12 L to 19 L, what is the final temperature? TV -1 = constant T,V For diatomic gas, C V = 5/2 R For diatomic gas, C P = 7/2 R C P /C V = 1.4000 for diatomic gas T i V i -1 = T f V f -1 Solving for T f gives 258 K 20
Example: 30 moles of a monatomic ideal gas at 1 atmosphere pressure expands adiabatically from an initial volume of 1.5 m 3 to 3.0 m 3. PV = constant P,V P T f Initial state Final state T i V a. What is the final pressure? C P (Note this is an approximation to [5/3]) C V = = 1.667 constant = P i V i = 1.99 x 10 5 Pa m 5 P f V f = 1.99 x 10 5 Pa m 5 P f V f nr P f = 3.19 x 10 4 Pa T f = =383 K P i = 1.01 x 10 5 Pa T i =607 K 21
b. What is the work done on the gas? Won PdV PV constant C = 1.99 x 10 5 W on C dv V Note that P must be inside the integral! C V 1 V 1 V 1 3 1.667 1 0.667 1 2 f V i 3 0.667 0.667 on f i W C V V 2 3C 3C 0.4807 0.7631 0.2824 2 2 84,350J 84.4kJ (also equals change in internal energy) 22
The work can also be calculated from W adiabatic P V PV f f i i 1 4 5 3.1910 31.01310 1.5 84,375 J γ-1 C. Is TV really a constant? TV T V 1 1 i i f f 1.667 1 0.667 0.667 607 1.5 383 3 6071.310 3832.080 795.2 796.7 To do better, use 2/3=0.666666667 instead of 0.667 throughout the calculation. 23
In summary, for the process under consideration, we have: 1 atm PV = constant 383 K 607 K 0.32 atm 1.5 m 3 3.0 m 3 Area under curve 84,375 J 24