The First Law of Thermodynamics

Transcription

1 Chapter 9 The First Law of Thermodynamics Topics for Chapter 9 I. First Law of Thermodynamics Internal energy, concept of state variables Difference between Work and Heat II. Examine various types of thermodynamic processes: Constant volume Constant pressure PowerPoint Lectures for University Physics, Twelfth Edition Hugh D. Young and Roger A. Freedman Lectures by James Pazun Modified by P. Lam 8_6_00 Constant temperature Zero heat transfer - adiabatic process First Law of Thermodynamics st Law of Thermodynamics is a statement about conservation of energy and it categorizes the method of energy transfer into two basic forms: work (W) and heat (Q). The internal energy of a system (U) (for a container of ideal gas, U =kinetic energy of the molecules) can be changed by transferring heat to and from the environment and/or performing work on or by the environment. U f!u i = "U = Q - W Positive Q # heat input to the system from the environment Negative Q # heat output from the system to the environment Positive W # work done by the system on the environment Negative W # work done on the system by the environment Note : The combined energy of the system and the environment is conserved; energy merely transferred to and from system and environment. Internal energy, concept of state variables Internal energy = energy of the system with respect to the center of mass reference; we are not concerned with the situation where the entire system is moving. Example: The internal energy of n moles of monatomic ideal gas is the kinetic energy of the gas with respect to the center of mass reference frame and it is U=3/nRT=3/PV. U is called a state variable because once the state of the system is specified, such as the number of moles and the temperature or pressure and volume, U has an unique value, that is U is function U(n,T) or U(P,V). The change of a state variable does not depend on how the system goes from the initial to the final state. For example, you can add heat or do work to change the system s from T i to T f, then ΔU=3/nRT f -3/nRT i no matter what methods was used.

2 Work = pressure x change of volume Consider the work done by the system (e.g. a gas) on the environment. W = Fdx = PAdx = PdV If dx > 0! dv > 0! volume expands! system does positive work on the enviroment According to the first Law : load "U = Q - W! positive work on the evironment lowers the internal energy of the system If dx < 0! compression, then W < 0 and the internal energy increases. (see picture to the right) Work done by a varying pressure is given by integration V f W =! P(V )dv V i Note : ()The direction of the path (indicated by an arrow) determines whether W is positive or negative. () W depends on the path going from initial to final state (see next slide) " W is NOT a state variable; there is NO such thing as #W = W f $W i Work done depends on the path. Shown below 3 different paths going from state to state ; the work done are different. What is energy transfer by heat? Imagine heating water with a burner. Energy is transferred from the fast moving molecules in the flame to the slow moving molecules in the pot and then to the slow moving molecules in the water by means of random collisions. At the microscopic level, the molecules in the flame are doing work (force x displacement) on the molecules in the pot which in turns are doing work on the water molecules. At this level, there no different between this method of energy transfer and the one described previously by pushing a piston. However, on the macroscopic level, work done by pushing a piston is done uniformly by changing the volume while heat can be added/removed without changing the volume. The first Law of thermodynamics treat work and heat on equal footing. The second Law of thermodynamics will treat heat differently than work.

3 II. Examine various types of thermodynamic processes Constant volume - isochoric process Process from to 4 is an isochoric process. = 3x0 5 N m,p =x0 5 N m =0! m 3,n = moles Evaluate "U,W,Q, and "T for the process to 4. Since there is no volume change! W = " PdV = 0 that is why it is called "isochoric".! #U = Q We won't be able to say anything further unless we know something about the system. Let the system be n moles of monatomic ideal gas, then U = 3 nrt = 3 PV, #U = U f = 3 P V $ 3 P = negative = $3kJ (OR use #U = U f = 3 nrt 4 $ 3 nrt = negative;t 4 = 60K,T =80K)! In this process, heat is released at constant volume to lower the pressure of the gas (lower the temperature of the gas) II. Examine various types of thermodynamic processes Constant pressure - isobaric process Process from to 3 is an isobaric process. W =! PdV = P ( - ) = +6kJ " #U = Q - P ( - ) We won't be able to say anything further unless we know something about the system. Let the system be n moles of monatomic ideal gas, then U = 3 nrt = 3 PV, = 3x0 N 5 m,p N #U = U f = 3 PV $ 3 PV = 3 =x05 m P (V $V ) = +9kJ =0! m 3, Evaluate "U,W,Q,and "T (OR use #U = U f = 3 nr(t $ T ) 3 " Q = #U +W = 5 P ( $ ) = 5 nr(t 3 $ T ) =5kJ In this process, heat is must be added to keep the pressure constant while doing work on the environment. II. Examine various types of thermodynamic processes Constant temperature - isothermal process Let process from to be an isothermal process. In this case, we won't be able to say anything at all unless we know something about the system (such as P vs V curve for T = constant). Let the system be n moles of monatomic ideal gas, then P = nrt V, T = constant = T = T =80K W =! Pd =! nrt V d = nrt ln = 3ln3 (kj) = 3x0 N 5 m,p =x0 N 5 m U = 3 nrt " #U = 3 nr#t = 0 =0! m 3, Evaluate "U,W,Q,and "T (Note : isothermal strictly means constant T; for ideal gas it also implies constant U, not true for real gases) Q = #U +W = W In this process, heat is must added to keep the temperature constant while doing work on the environment. Aside - Re-enforce the concept of state variables N = 3x0 5 m,p N =x05 m =0! m 3, of monatomic ideal gas Evaluate "U,W,Q,and "T for the path - > 3- > and compare with the isothermal process - > 3

4 II. Examine various types of thermodyanmic processes No heat transfer (Q=0) - adiabatic process Let process from to be an adiabtic process. All we can say is!u = Q - W = -W We won't be able to say anything further unless we know something about the system (such as P vs V curve when Q = 0). Let the system be n moles of monatomic ideal gas, then P = nrt V, however T " constant, T =80K nrt W = # Pd = V dv V # " nrt ln Let P = 3x0 N 5 m,p =??? because T is also a function of V for this path. =0! m 3, Evaluate "U,W,Q,and "T Since W = positive (see direction of arrow => expansion) $!U = Q - W = -W = negative $ T decreases along the path. For monatomic ideal gas, we can use!u = 3 nr ( T % T ) = %W but need to know T (see next slide) Adiabatic PV curve for monatomic ideal gas In an adiabatic process, no heat is transferred from system and Derive the adiabatic PV curve for an ideal gas. surroundings. = P a = 3x0 5 N m,p = P b =??? = V a =0! m 3, = V b Evaluate "U,W,Q,and "T Use differential form of the st Law : du = dq - dw = -dw = -PdV For ideal gas : U = 3 nrt = 3 PV! du = 3 PdV + 3 VdP! 3 PdV + 3 VdP = "PdV! 5 PdV = " 3 VdP! 5 dv V = " 3 dp P! 5 lnv V = " 3 lnp P P! 5 ln V = " 3 ln P P! V 5 / % ( = P "3 / % ( = P 3 / % ( $ ' $ P ' $ P ' # V! P = P & # V & % ( in particular : P = P % ( $ ' = 3x0 5 5/3 N % $ 3 ( ' m Use P = nrt / 3 / 3 V! T = T # & # V % (! T = T & % ( $ ' =80 /3 % $ 3 ( K ' V # V W = ) PdV = & 5 / ) P % ( dv = P V 3 " 3 V * V - " / 3 +,. / = 3 * P " / 3 -, % ( / +, $ './ = 3 P " 3 P = 3 nr ( T " T ) = "0U Compare the four processes on a PV diagram Notice the subtle differences for each curve in Figure 9.6. Cyclic process - complete cycle Cyclic process :! 3!! 4! No matter what the substance is, "U complete cycle = 0 because it comes back to the same state "U = U # U 4

5 Relating heat capacities at constant volume and pressure Compare molar heat capacities, c V and c P, for monatomic ideal gas Constant volume (W = 0) : Q =!U = 3 nr!t " c V = 3 R Constant pressure : W = -P!V " Q =!U +W =!U +P!V PV = nrt " P!V = nr!t (for constant P) " Q = 3 nr!t + nr!t = n(c + R)!T v " c p = c v + R It requires more heat to raise the temperature at constant pressure vs constant volume because some of the heat is spent in doing work, expanding the volume. Measuring heat capacities Heat capacities may be measured at constant volume in a fairly complex process using a bomb calorimeter. Heat capacities may be measured at constant pressure using equipment as simple as a coffee cup. For solids, c p # c v, because the volume does not expand as easily as a gas. Nonetheless c p is always greater than c v. Heat capacities tabulated for selected gasses J c p = c v + R; R ~ 8.3 mol K monatomic : c v = 3 R! c p = 5 R! c p c v = 5 3 ".67 diatomic : c v = 5 R (without vibration)! c p = 7 R! c p c v = 7 5 ".4 5