First Law of Thermodynamics

Transcription

1 First Law of Thermodynamics E int = Q + W other state variables E int is a state variable, so only depends on condition (P, V, T, ) of system. Therefore, E int only depends on initial and final states of the system. Q and W, however, depend on the process.

2 E int = Q + W We will mostly be concerned with (nonmagnetic, nonelectric) liquids and gases, for which W = - P dv (if the pressure stays well defined during the process: quasi-static) Types of Processes: 1) Isothermal: System is in contact with an ( infinite ) energy reservoir (also called a heat bath ) that keeps its temperature constant. 2) Isobaric: P = constant: W = -P V 3) Isovolumetric: V = constant: W = 0, E int = Q. 4) Adiabatic Process: Q = 0. Either: a) in perfectly insulated container b) occurs rapidly (and perhaps not quasi-static) so heat does not have enough time to flow in/out. (e.g. this is what happens during power and compression strokes in engines) Example: Adiabatic Free Expansion: Gas rushing to fill a volume. Not quasi-static and pressure not defined during the process. Q = 0, W = 0 E int = 0. Since Q depends on process, heat capacity C dq/dt also depends on process: Molar specific heats: C P (1/n) dq/dt when P = constant ; C V (1/n) dq/dt when V = constant. [For liquids and solids, C P C V, but they can be quite different for gases.]

3 Problem: Ethanol, with an initial volume of V 0 = m 3, is in a cylinder with a piston at a constant pressure of 2 atm. The ethanol is heated from 20 o C to 35 o C. What is the change in its internal energy? Ethanol: c = 2400 J/kg o C [or C V C P = (2400 J/kg o C) (0.046 kg/mol)] β = 1.12 x 10-4 / o C ρ = 790 kg/m 3 Q = mc T [= nc P T] m = m kg/m 3 = 3.16 kg [or n=3.16 kg/(0.046 kg/mol)] Q = (3.16 kg)(2400 J/kg o C) (15 o C) Q = 1.14 x 10 5 J F = PA W = -P V = -P βv 0 T W = - (2 x 1.01 x 10 5 N/m 2 ) (1.12 x 10-4 / o C) (0.004 m 3 ) (15 o C) W = J [W<0 because ethanol expands, doing work on surroundings] E int = Q + W Q = 1.14 x 10 5 J [ Q >> W, as is usually case for liquids and solids, since their volumes are almost constant i.e. they are very incompressible]

4 Cycle System returns to initial T,V,P,E int (so E int = 0), but there may be net Q and net W. For cycle with steps j : E int (cycle) = E int (j) = Q j + W j = 0, but may have Q j = - W j 0. For clockwise (quasi-static) cycle, net W = W j = - area inside. For counterclockwise (quasi-static) cycle, net W = W j = + area inside.

5 Some Properties of an Ideal Gas 1) Equation of State: P = nrt/v (definition of Kelvin temperature scale) 2) E int n (i.e. it is extensive proportional to amount of gas) 3) E int / n only depends on T [= (PV)/(Rn)], so if T is constant, E int is also constant. (i.e. E int does not separately depend on V or p.) discovered by Joule 4) C P C V = R (Derived in Ch. 21.) 5) For a monoatomic ideal gas (e.g. He, Ne, Ar, but not N 2, O 2, H 2, CO 2, ), C V = 3/2 R, C P = 5/2 R (Derived in Ch. 21) Adiabatic Free Expansion: Since E int = 0 T = 0. Quasi-Static Isothermal Process: P = nrt/v, W = - nrtdv/v = -nrt dv/v = -nrt ln(v f /V i ) Since T = constant, E int = 0, so Q = - W = +nrt ln(v f /V i )

6 n moles of helium (monoatomic ideal gas) i Problem: Find Q and W for each step and check that E int = Q + W = 0.

7 n moles of helium (monoatomic ideal gas) i W AB = 0 (because V = 0) Q AB = nc V (T B -T A ) = 3/2 nr (3P i V i /nr P i V i /nr) = 3P i V i W BC = -3P i (3V i -V i ) = -6P i V i Q BC = nc P (T C -T B ) = 5/2 nr (9P i V i -3P i V i )/nr = 15 P i V i W CD = 0 Q CD = nc V (T D -T C ) = 3/2 nr (3P i V i /nr 9P i V i /nr) = -9P i V i W DA = P i (3V i -V i ) = 2P i V i Q DA = nc P (T A -T D ) = 5/2 nr (P i V i -3P i V i )/nr = -5 P i V i W = 0-6PiVi+0+2PiVi } W = -4 P i V i [ W<0: net work done by gas] Q = ( )P i V i Q = +4 P i V i [ Q > 0: net heat into gas] E int (cycle) = Q + W = 0

8 n moles of helium 3P 0 A isotherm Pressure P 0 B C V 0 3 V 0 Volume

9 n moles of helium 3P 0 A isotherm Pressure P 0 B C W AB = 0 Q AB = nc V (T B T A ) = 3/2 nr (P 0 V 0 3P 0 V 0 )/nr = -3P 0 V 0 W BC = -P 0 (3V 0 V 0 ) = -2 P 0 V 0 Q BC = nc P (T C T B ) = 5/2 nr (3P 0 V 0 P 0 V 0 ) = 5 P 0 V 0 CA: isotherm W CA = -nrt ln[(v 0 /3V 0 ] = 3P 0 V 0 ln(3) But E int (C A) = 0 since isothermal Q CA = -W CA = -3P 0 V 0 ln(3) V 0 3 V 0 Volume } W = 0-2P0V0 +3P0V0 ln(3) W = P 0 V 0 [3ln(3) -2] [ W>0: net work done on gas] Q = ( ln(3))p 0 V 0 Q = [2-3 ln(3)] P 0 V 0 [ Q < 0: net heat out of gas] E int (cycle) = Q + W = 0

10 Problem: Find Q and W for each step.

11 n moles of helium 3P 0 W BC = 0 Q BC = nc V (T C -T B ) = 3/2 nr (3P 0 V 0-9P 0 V 0 )/nr = - 9P 0 V 0 W CA = -P 0 (V 0-3V 0 ) = + 2P 0 V 0 Q CA = nc P (T A -T C ) = 5/2 nr (P 0 V 0 3P 0 V 0 ) / nr = -5P 0 V 0 P 0 W AB = - pdv = - area under AB = -[½ (2P 0 *2V 0 ) + P 0 *2V 0 ] = - 4P 0 V 0 Q AB =?? V 0 3V 0

12 n moles of helium 3P 0 P 0 V 0 3V 0 To find Q AB, use E int (cycle) = Q + W = 0 Q AB = (Q BC + Q CA + W AB + W BC + W CA ) Q AB = - ( ) P 0 V 0 Q AB = 16 P 0 V 0

13 8 moles of Argon (a monoatomic ideal gas) undergo the cycle below, in which step CD is an isotherm and AB is adiabatic. How much work is done in step AB?

14 8 moles of Argon (a monoatomic ideal gas) undergo the cycle below, in which step CD is an isotherm and AB is adiabatic. How much work is done in step AB? E int = 0 = W AB + Q AB + W BC + Q BC + W CD + Q CD + W DA + Q DA [W CD = - Q CD ] 0 = W AB + 0 P B (V C V B )+ n C P (T C -T B ) P A (V A V D ) + nc P (T A T D ) [at any point: nc P T = 5/2 nr (PV/nR) = 5/2 PV] W AB = [(3.03 x 10 5 ) (0.31) -5/2 (3.03 x 10 5 ) (0.31) (1.01 x 10 5 )(1.00) + 5/2 (1.01 x 10 5 ) (1.00)] (N/m 2 ) (m 3 ) W AB = 3/2 ( x x 10 5 ) J = x 10 4 J

15 C, T T 0 Suppose an object, with heat capacity C is at temperature T, in thermal contact with surroundings, which are at temperature T 0. Q will flow into object (Q >0) if T < T 0 and Q will flow out (Q< 0) if T > T 0. dq = C dt dq/dt = C dt/dt

16 dq/dt = C dt/dt [Note: dq/dt > 0 if T < T 0 and dq/dt <0 if T > T 0.] Mechanisms of Heat Flow 1) Conduction 2) Radiation 3) Convection

17 Convection: Heating by matter transfer Suppose there is a temperature gradient in a gas or liquid in a gravitational field. If the density varies with temperature [e.g. ideal gas: n/v = P/RT, so density decreases with increasing T if P = constant ], then the more dense (usually the colder) material will sink (move in direction of gravitational force) and the less dense material will flow opposite the gravitational field. That is, the warmer material will flow to the colder regions and the colder material to the warmer regions warm air cool air g Earth

18 Water There will also be convection in water. For T > 4 o C, warm water is less dense and will rise. However, for 0 o C < T < 4 o C, cold water is less dense and will rise lakes freeze from the top.