# The first law of thermodynamics 10.2.1 Deduce an expression for the work involved in a volume change of a gas at constant pressure. 10.2.2 State the first.

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The first law of thermodynamics 10.2.1 Deduce an expression for the work involved in a volume change of a gas at constant pressure. 10.2.2 State the first law of thermodynamics. 10.2.3 Identify the first law of thermodynamics as a statement of energy conservation. 10.2.4 Describe the isochoric (isovolumetric), isobaric, isothermal, and adiabatic changes of state of an ideal gas. 10.2.5 Draw and annotate thermodynamic processes and cycles on P-V diagrams. 10.2.6 Calculate from a P-V diagram the work done in a thermodynamic cycle. 10.2.7 Solve problems involving state changes of a gas. Topic 10: Thermal physics 10.2 Processes

The first law of thermodynamics Deduce an expression for the work involved in a volume change of a gas at constant pressure.  Suppose we take a beaker that is filled with an ideal gas, and stopper it with a gas-tight cork and a weight, as shown.  The weight F causes a pressure in the gas having a value given by P = F/A, where A is the area of the cork in contact with the gas.  If we now heat up the gas it will expand against the cork, pushing it upward:  The dashedred box shows the change in volume ∆V.  Note that ∆V = Ax. Topic 10: Thermal physics 10.2 Processes x P is constant. Why? ∆V∆V A

The first law of thermodynamics Deduce an expression for the work involved in a volume change of a gas at constant pressure.  From the previous slide: P = F/A  F = PA and ∆V = Ax.  The work W done by the gas is just the force F it exerts on the weighted cork times the distance x it moves the cork. Thus W = Fx = PAx = P∆V. FYI  If ∆V > 0 (gas expands) then W > 0.  If ∆V < 0 (gas contracts) then W < 0. Topic 10: Thermal physics 10.2 Processes x ∆V∆V A F Work done by expanding gas (constant P) W = P∆V

The first law of thermodynamics State the first law of thermodynamics.  Consider the previous “system” containing gas, and heat, and providing mechanical work.  Just as we have done before, we will use Q to represent a quantity of heat. If heat is added to our system (as it was) then Q > 0. If heat is removed from our system then Q < 0.  If the gas expands then W > 0. If the gas contracts then W < 0.  Finally, we define the change in internal energy of the system as ∆U.  The first law of thermodynamics relates Q, W and ∆U as follows: Topic 10: Thermal physics 10.2 Processes first law of thermodynamics Q = ∆U + W

The first law of thermodynamics Identify the first law of thermodynamics as a statement of energy conservation.  What the first law shows is that when heat energy Q is added to a system, some (or all of it) may be used to change the internal energy ∆U of the system, and some (or all of it) may be used to provide mechanical work W. Topic 10: Thermal physics 10.2 Processes first law of thermodynamics Q = ∆U + W FYI  Sometimes the first law is expressed ∆U = Q – W.  In this form we see that there are two ways to change the internal energy of a system: (1)By passing heat Q through the system boundary. (2)By doing work W through the system boundary.

The first law of thermodynamics Identify the first law of thermodynamics as a statement of energy conservation.  In words, the first law says “Heat added to a closed system can change its internal energy and cause it to do work on its environment.”  This is a statement of the conservation of energy. All of the energy added to the system is accounted for. Topic 10: Thermal physics 10.2 Processes first law of thermodynamics Q = ∆U + W FYI  Recall that ∆U consists of both ∆E P (manifested in phase change) and ∆E K (temperature change) of the substance. Lack of internal forces in an ideal gas means that ∆E P = 0.

The first law of thermodynamics Describe the isochoric (isovolumetric), isobaric, isothermal, and adiabatic changes of state of an ideal gas.  For an ideal gas, the equation of state (PV = nRT) tells us all of the important things about a gas through the state variables P,V, n, and T.  When we add (or remove) heat Q from a closed system containing an ideal gas, or have it do work W on the external world, its state variables may change.  We call changing the state of a system a process.  A process in which the state of a gas is changed without changing its volume is called isochoric. Topic 10: Thermal physics 10.2 Processes FYI  Don’t confuse “state” with “phase” in Topic 10.

The first law of thermodynamics Describe the isochoric (isovolumetric), isobaric, isothermal, and adiabatic changes of state of an ideal gas.  A process in which the state of a gas is changed without changing its volume is called isochoric.  We have already seen an isochoric process when we studied the concept of absolute zero: Topic 10: Thermal physics 10.2 Processes 0 10 20 30 p T (°C) -300 -200-100 0 100 200 300 During an isochoric process the temperature and the pressure change. NOT the volume.

The first law of thermodynamics Describe the isochoric (isovolumetric), isobaric, isothermal, and adiabatic changes of state of an ideal gas.  A process in which the state of a gas is changed without changing its volume is called isochoric. Topic 10: Thermal physics 10.2 Processes PRACTICE: Show that the first law of thermodynamics reduces to Q = ∆U for an isochoric process. SOLUTION:  Recall that the work done by a gas is given by W = P∆V.  Isochoric means ∆V = 0. Thus W = 0.  The first law, Q = ∆U + W, thus reduces to Q = ∆U.

The first law of thermodynamics Describe the isochoric (isovolumetric), isobaric, isothermal, and adiabatic changes of state of an ideal gas.  A process in which the state of a gas is changed without changing its volume is called isochoric. Topic 10: Thermal physics 10.2 Processes EXAMPLE: Show that for an isolated ideal gas P  T during an isochoric process. SOLUTION: Use PV = nRT. Then P = (nR/V)T  Isolated means n is constant (no gas is added to or lost from the system).  Then n and V are constant (as is R). Thus P = (nR/V)T = (CONST)T P  T. (Isochoric)

The first law of thermodynamics Describe the isochoric (isovolumetric), isobaric, isothermal, and adiabatic changes of state of an ideal gas.  A process in which the state of a gas is changed without changing its pressure is called isobaric.  We have already seen such a process, when we derived the formula for the work done by a gas:  For an isobaric process the first law can therefore be written Q = ∆U + P∆V. Topic 10: Thermal physics 10.2 Processes x ∆V∆V A Work done by expanding gas (constant P) W = P∆V

The first law of thermodynamics Describe the isochoric (isovolumetric), isobaric, isothermal, and adiabatic changes of state of an ideal gas.  A process in which the state of a gas is changed without changing its pressure is called isobaric. Topic 10: Thermal physics 10.2 Processes EXAMPLE: Show that for an isolated ideal gas V  T during an isobaric process. SOLUTION: Use PV = nRT. Then V = (nR/P)T  Isolated means n is constant (no gas is added to or lost from the system).  Then n and P are constant (as is R). Thus V = (nR/P)T = (CONST)T V  T. (Isobaric)

The first law of thermodynamics Describe the isochoric (isovolumetric), isobaric, isothermal, and adiabatic changes of state of an ideal gas.  A process in which the state of a gas is changed without changing its pressure is called isobaric. Topic 10: Thermal physics 10.2 Processes PRACTICE: Show that for an isolated ideal gas W = nR∆T during an isobaric process. SOLUTION:  From PV = nRT we can write (if n and P are constant) P∆V = nR∆T.  Recall W = P∆V.  Thus W = nR∆T. (Isobaric)

The first law of thermodynamics Describe the isochoric (isovolumetric), isobaric, isothermal, and adiabatic changes of state of an ideal gas.  If the state of a gas is changed without changing its temperature the process is called isothermal. Topic 10: Thermal physics 10.2 Processes EXAMPLE: A graduated syringe which is filled with air is placed in an ice bath and allowed to reach the temperature of the water. Demonstrate that P 1 V 1 = P 2 V 2. SOLUTION:  Record initial states after a wait: P 1 = 15, V 1 = 10, and T 1 = 0ºC.  Record final states after a wait: P 2 = 30, V 2 = 5, and T 2 = 0ºC. P 1 V 1 = 15(10) = 150 = 30(5) = P 2 V 2. 0 10 20 30 How do we know that the process is isothermal? Why do we wait before recording our values?

The first law of thermodynamics Describe the isochoric (isovolumetric), isobaric, isothermal, and adiabatic changes of state of an ideal gas.  If the state of a gas is changed without changing its temperature the process is called isothermal. Topic 10: Thermal physics 10.2 Processes PRACTICE: Show that for an isolated ideal gas P 1 V 1 = P 2 V 2 during an isothermal process. SOLUTION:  From PV = nRT we can write (if n and T are constant) P 1 V 1 = nRT P 2 V 2 = nRT.  Thus P 1 V 1 = nRT = P 2 V 2. (Isothermal)

The first law of thermodynamics Describe the isochoric (isovolumetric), isobaric, isothermal, and adiabatic changes of state of an ideal gas.  If the state of a gas is changed without adding or losing heat the process is called adiabatic. Topic 10: Thermal physics 10.2 Processes PRACTICE: Show that for an isolated ideal gas W = -∆U during an adiabatic process. SOLUTION:  From Q = ∆U + W we can write (if n is constant and Q is zero) Q = ∆U + W 0 = ∆U + W W = -∆U. FYI  We accomplish this by insulating the container.

The first law of thermodynamics Draw and annotate thermodynamic processes and cycles on P-V diagrams.  Perhaps you have enjoyed the pleasures of analytic geometry and the graphing of surfaces in 3D.  The three variables of a surface are x, y, and z, and we can describe any surface using the "state" variables x, y, and z:  The equation "of state" of a sphere is x 2 + y 2 + z 2 = r 2, where r is the radius of the sphere: Topic 10: Thermal physics 10.2 Processes FYI  We “built” the 3D sphere with layers of 2D circles.  We have transformed a 3D surface into a stack of 2D surfaces. y z x

The first law of thermodynamics Draw and annotate thermodynamic processes and cycles on P-V diagrams.  The three state variables (if n is kept constant) of a gas are analogous.  We can plot the three variables P, V, and T on mutually perpen- dicular axes like this:  We have made layers in T. Thus each layer has a single temperature. Topic 10: Thermal physics 10.2 Processes FYI  Each layer is an isotherm.  The 3D graph (below) can then be redrawn in its simpler 2D form (above) without loss of information. V P T V P i s o t h e r m s T 1 T 2 T 3 T 4

The first law of thermodynamics Draw and annotate thermodynamic processes and cycles on P-V diagrams.  A thermodynamic process involves moving from one state to another state. This could involve changing any or even all of the state variables (P, V, or T). Topic 10: Thermal physics 10.2 Processes EXAMPLE: In the P-V graph shown, identify each process type as ISOBARIC, ISOTHERMAL, OR ISOVOLUMETRIC (isochoric). SOLUTION:  A  B is isothermal (constant T).  B  C is isobaric (constant P).  C  A is isochoric (constant V). FYI  The purple line could be an adiabatic process. P i s o t h e r m i s o t h e r m A B C V

The first law of thermodynamics Draw and annotate thermodynamic processes and cycles on P-V diagrams.  A thermodynamic cycle is a set of processes which ultimately return a gas to its original state. Topic 10: Thermal physics 10.2 Processes EXAMPLE: A fixed quantity of a gas undergoes a cycle by changing between the following three states: State A: (P = 2 Pa, V = 10 m 3 ) State B: (P = 8 Pa, V = 10 m 3 ) State C: (P = 8 Pa, V = 25 m 3 ) Each process is a straight line, and the cycle goes like this: A  B  C  A. Sketch the complete cycle on a P-V diagram. SOLUTION:  Scale your axes and plot your points… P V 2 8 1025 A B C

The first law of thermodynamics Calculate from a P-V diagram the work done in a thermodynamic cycle.  A thermodynamic cycle is a set of processes which ultimately return a gas to its original state. Topic 10: Thermal physics 10.2 Processes EXAMPLE: A fixed quantity of a gas undergoes the cycle shown here (from the last example): (a) Find the work done during the process A  B. (b) Find the work done during the process B  C. SOLUTION: Use W = P∆V. (a) From A to B: ∆V = 0. Thus the W = 0. (b) From B to C: ∆V = 25 – 10 = 15; P = 8. Thus W = P∆V = 8(15) = 120 J. P V 2 8 1025 A B C

EXAMPLE: A fixed quantity of a gas undergoes the cycle shown here (from the last example): (c) Find the work done during the process C  A. SOLUTION: (c) Use W = Area under the P-V diagram.  Observe that ∆V is negative when going from C (V = 25) to A (V = 10).  Observe that P is NOT constant so W  P∆V.  W = Area = -[ (2)(15) + (1/2)(6)(15) ] = -75 J. The first law of thermodynamics Calculate from a P-V diagram the work done in a thermodynamic cycle.  A thermodynamic cycle is a set of processes which ultimately return a gas to its original state. Topic 10: Thermal physics 10.2 Processes P V 2 8 1025 A B C

EXAMPLE: A fixed quantity of a gas undergoes the cycle shown here (from the last example): (d) Find the work done during the cycle A  B  C  A. SOLUTION: (d) Just total up the work done in each process.  W A  B = 0 J.  W B  C = + 120 J.  W C  A = -75 J.  W cycle = 0 + 120 – 75 = + 45 J. FYI  Work is done on the external environment during each cycle. The first law of thermodynamics Calculate from a P-V diagram the work done in a thermodynamic cycle.  A thermodynamic cycle is a set of processes which ultimately return a gas to its original state. Topic 10: Thermal physics 10.2 Processes P V 2 8 1025 A B C

PRACTICE: Find the total work done if the previous cycle is reversed. SOLUTION:  We want A  C  B  A.  W A  C = Area = +[ (2)(15) + (1/2)(6)(15) ] = + 75 J.  W C  B = P∆V = 8(10–25) = - 120 J.  W B  A = 0 J (since ∆V = 0).  W cycle = 75 - 120 + 0 = - 45 J. FYI  Reversing the cycle reverses the sign of the work. The first law of thermodynamics Calculate from a P-V diagram the work done in a thermodynamic cycle.  A thermodynamic cycle is a set of processes which ultimately return a gas to its original state. Topic 10: Thermal physics 10.2 Processes P V 2 8 1025 A B C

The first law of thermodynamics Calculate from a P-V diagram the work done in a thermodynamic cycle.  A heat engine is any device which converts heat energy into mechanical work.  To make a heat engine we need a source of heat energy (coal, gas, etc.) and a working fluid which undergoes thermodynamic change of state causing work do be done on the external environment.  Common working fluids are water (made into steam) and petrol-air mixtures (ignited).  The energy flow diagram of a heat engine is shown here: Topic 10: Thermal physics 10.2 Processes hot reservoir at T H cold reservoir at T L engine QHQH QLQL W

EXAMPLE: An internal combustion engine is an example of a heat engine that does work on the environment. A four-stroke engine is animated here. http://www.animatedengines.com/otto.html http://chemcollective.org/activities/simulations/ engine The first law of thermodynamics Calculate from a P-V diagram the work done in a thermodynamic cycle. Topic 10: Thermal physics 10.2 Processes

EXAMPLE: From the animation of the second link on the previous slide, show the direction of each process with an arrow. SOLUTION:  View the animation. Or reason that POSITIVE work must be done by the cycle. FYI  Remember “work done BY system” is positive, and “work done ON system” is negative. The first law of thermodynamics Calculate from a P-V diagram the work done in a thermodynamic cycle. Topic 10: Thermal physics 10.2 Processes

EXAMPLE: From the same animation, label the COMPRESSION, POWER, EXHAUST, AND INTAKE STROKES. Which stroke has a positive area under it? Which has a negative area? SOLUTION:  A STROKE involves a change in volume.  Since ∆V > 0 for the power stroke it has a positive area.  The compression stroke has a negative area. The first law of thermodynamics Calculate from a P-V diagram the work done in a thermodynamic cycle. Topic 10: Thermal physics 10.2 Processes COMPRESSION STROKE POWER STROKE EXHAUST STROKE INTAKE STROKE

EXAMPLE: From the same animation, show that the work done in a cycle is positive. SOLUTION:  The work done during the power stroke is positive (sketched in red).  The work done during the compression stroke is negative (sketched in purple).  There is more positive than negative. Thus W > 0.  W = area between graphs. The first law of thermodynamics Calculate from a P-V diagram the work done in a thermodynamic cycle. Topic 10: Thermal physics 10.2 Processes COMPRESSION STROKE POWER STROKE EXHAUST STROKE INTAKE STROKE

The first law of thermodynamics Calculate from a P-V diagram the work done in a thermodynamic cycle.  A heat pump is any device which can move heat from a low temperature reservoir to a high temperature reservoir.  A refrigerator is an example of a heat pump.  A common working fluid is Freon.  The Freon vaporizes inside the fridge removing Q L = L V.  The Freon condenses outside the fridge releasing Q H = L f. Topic 10: Thermal physics 10.2 Processes hot reservoir at T H cold reservoir at T L pump QHQH QLQL W FYI  The compressor does the work ON the system.

The first law of thermodynamics Solve problems involving state changes of a gas. Topic 10: Thermal physics 10.2 Processes  ”Thermally insulated” means that heat can not enter or leave the system.  Thus Q = 0.  This is the definition of an adiabatic process.

The first law of thermodynamics Solve problems involving state changes of a gas. Topic 10: Thermal physics 10.2 Processes  W = P∆V = 1  10 5 (-3) = -3  10 6 J.  Or you can just find the area under the process.

The first law of thermodynamics Solve problems involving state changes of a gas. Topic 10: Thermal physics 10.2 Processes  Adiabatic (and insulated) both mean that Q = 0.  The first law says Q = ∆U + W. Then  ∆V < 0 means W < 0 (-). 0 = ∆U + (-)  ∆U > 0.

The first law of thermodynamics Solve problems involving state changes of a gas. Topic 10: Thermal physics 10.2 Processes  Isothermal means T is constant.  PV = nRT then becomes PV = CONST.  At A: PV = 2  5 = 10.  At B: PV = 5  2 = 10.  At C: PV = 6.8  2 = 13.6.  Thus process AB is the isothermal one.

The first law of thermodynamics Solve problems involving state changes of a gas. Topic 10: Thermal physics 10.2 Processes  The difference in work is the area between the graphs.

The first law of thermodynamics Solve problems involving state changes of a gas. Topic 10: Thermal physics 10.2 Processes  This can only be an estimate.  A tally of small rectangles should suffice. 18 16 14 12 10 9 9 8 8 7 7 6 5 5 4 4 19  There are about 170 of them.  Each rectangle has a work value given by W = PV = (0.1  10 5 )(0.1  10 -3 ) = 1 J.  Thus the difference in work is 170 J.

The first law of thermodynamics Solve problems involving state changes of a gas. Topic 10: Thermal physics 10.2 Processes  The first law says Q = ∆U + W.  Adiabatic means Q = 0. ∆U = -W.  AC is the adiabatic compression.  Since W < 0 during the compression of a gas, ∆U = -W > 0. Thus the temperature increases.

The first law of thermodynamics Solve problems involving state changes of a gas. Topic 10: Thermal physics 10.2 Processes  The gas only does external work when it expands.  QR and SP are isochoric (isovolumetric) so W = 0.  PQ is a compression so W < 0.  RS is an expansion so W > 0.

The first law of thermodynamics Solve problems involving state changes of a gas. Topic 10: Thermal physics 10.2 Processes  NOT adiabatic since Q  0 (Q = 8  10 3 J).  From PV = nRT we see that P i V i = 1.2  10 5 (0.05) = 6000 J = nRT i. P f V f = 1.2  10 5 (0.10) = 12000 J = nRT f.  Thus T changes and the process is NOT isothermal.  So the process is NEITHER.

The first law of thermodynamics Solve problems involving state changes of a gas. Topic 10: Thermal physics 10.2 Processes  From W = P∆V we see that W = 1.2  10 5 (0.10 - 0.05) = 6.0  10 3 J.

The first law of thermodynamics Solve problems involving state changes of a gas. Topic 10: Thermal physics 10.2 Processes  From the first law: Q = ∆U + W, or 8.0  10 3 = ∆U + 6.0  10 3 ∆U = 2.0  10 3 J.

The first law of thermodynamics Solve problems involving state changes of a gas. Topic 10: Thermal physics 10.2 Processes  Work done ON a gas is a compression.  Only process PQ has a decreasing volume.

The first law of thermodynamics Solve problems involving state changes of a gas. Topic 10: Thermal physics 10.2 Processes  For an ideal gas, ISOTHERMAL means ∆U = 0.  From the first law of thermodynamics we have Q = ∆U + W Q = 0 + W Q = W = 2500 J.

The first law of thermodynamics Solve problems involving state changes of a gas. Topic 10: Thermal physics 10.2 Processes  From the first law we have q = ∆U + w.  Keep T constant with ICE BATH.  Keep V constant with FIXED CONTAINER.  Keep q zero with INSULATION.

The first law of thermodynamics Solve problems involving state changes of a gas. Topic 10: Thermal physics 10.2 Processes  A heat pump transfers heat from a cold reservoir to a hot one. QHQH QLQL  Since this is not the natural direction of flow, work must be done ON the system to make this happen. W

The first law of thermodynamics Solve problems involving state changes of a gas. Topic 10: Thermal physics 10.2 Processes  It is clear from the diagram that both A  B and C  D are isobaric (P = CONST).  From C  D evaporation occurs. During phase changes ∆T = CONST (isothermal).  From A  B compression occurs. During phase changes ∆T = CONST (isothermal).

The first law of thermodynamics Solve problems involving state changes of a gas. Topic 10: Thermal physics 10.2 Processes  Wherever W > 0 heat is removed from the cold reservoir. Thus B  C and C  D.  Wherever W < 0 heat is forced into the hot reservoir. Thus D  A and A  B.

The first law of thermodynamics Solve problems involving state changes of a gas. Topic 10: Thermal physics 10.2 Processes  From the diagram E in = W + Q c. E out = Q H.  From E in = E out we have W + Q C = Q H.

The first law of thermodynamics Solve problems involving state changes of a gas. Topic 10: Thermal physics 10.2 Processes  A refrigerator is a heat pump. Work must be done ON it. Thus W < 0.  Area inside curves must be NEGATIVE.  P = CONST A A B  Wherever W > 0 heat is removed from the cold reservoir. Thus where ∆V > 0. B

The first law of thermodynamics Solve problems involving state changes of a gas. Topic 10: Thermal physics 10.2 Processes  For each square: W = P∆V W = (1  10 5 )(0.1  10 -3 ) W = 10 J.  There are about 47 squares.  Thus W = 47(10 J) = 470 J (done ON substance).

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