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Thermodynamics Practice Questions

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Question 1 1 mole of ideal gas is brought to a final state F by one of three processes that have different initial states as shown in the figure. What is true for the temperature change between initial and final states? It’s the same for all processes. It’s the smallest for process 1. It’s the smallest for process 2. It’s the smallest for process 3. It’s the same for processes 1 and 2. V0 2V0 p0 2p0 3p0 1 2 3 F The temperature for gas at any point is given by: Since all of the three processes end at the same temperature, we need only look at the process that starts closest to: Process 2 actually begins at this temperature, so its change is 0

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Question 2 1 mole of ideal gas is brought to a final state F by one of three processes that have different initial states as shown in the figure. What is true for the work done by the gas? It’s positive for process 1 and 2, but negative for process 3. It’s the smallest for process 1. It’s the smallest for process 2. It’s the smallest for process 3. Zero work is done along process 3. V0 2V0 p0 2p0 3p0 1 2 3 F Work done by the gas is equal to the area under the curve

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Question 3 An ideal gas is brought from S -> F by three different paths: SRF, SF, STF. The temperature at S is the same as the temperature at F. Which of the following is true? Process SRF occurs at constant temperarure The work done by the gas along SRF is the same as the work done by the gas along STF. Net heat into the gas along STF is greater than the work done by the gas along this path. The change in the internal energy is the same for all three paths. Work done by the gas along STF is greater than the work done by the gas along SF. The internal energy of an ideal gas depends only on temperature The internal energy of an ideal gas depends only on temperature. If the initial and final temp are the same, there is no change in internal energy. U=0 for all three paths. F S T R V P W= - Q for each process, which eliminates C Area considerations eliminate B and E Temperature at R is greater than the temperature at S (gas law) eliminates A

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**Question 4 What is true for the process D -> A? U=0 Q>0.**

W= U>0 W= U<0 W= Q=0 V0 2V0 p0 2p0 C A B D Since the area (work) under D -> A is zero, this eliminates A and B E is eliminated because if Q = 0 and W=0, then U would have to be zero, thus no change in P could occur. The internal energy must have increased to double the pressure, therefore U>0.

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**Question 5 What is true for the two step process A -> B -> C?**

U=0 Q=0 U=0 Q>0 W= Q>0 W= Q<0 W> Q<0 V0 2V0 p0 2p0 C A B D Since the area (work) under A -> B -> C is not zero, this eliminates C and D A is eliminated because if U=0 and Q=0 then W must be 0, but it is not. The work done on the gas is negative (the gas does positive work when expanding), therefore Q>0, this eliminates D and E The ideal gas law tells us that the temperature does not change between A and C, so U=0. and since expanding Q>0.

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Question 6 200 J enters a Carnot engine from the hot reservoir, held at 400 K. During the entire engine cycle, 50 J of useful work is performed by the engine. What is the temperature of the cold reservoir? 100 K 200 K 300 K 250 K 150 K From the definition of efficiency: Since this is a Carnot engine, we have

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Question 7 For an ideal gas in a container with fixed volume and constant temperature, which of the following is true? Pressure results from molecular collisions with the walls. The molecules all have the same speed. The average kinetic energy is directly proportional to the temperature I, II, and III I and II only II and III only II only I and III only The distribution of speeds among the molecules, leading to an average value, but the molecules certainly do not all have the same speed.

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Question 8 1 mole of ideal gas is in a container of volume V, temperature T, and pressure P. If the volume is halved and the temperature doubled, the pressure will be: P 2P 4P ½ P ¼ P Since an ideal gas:

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Question 9 Two identical containers contain 1 mole each of two different monatomic ideal gases, gas A and gas B, with the mass of gas B 4 times the mass of gas A. Both gases are at the same temperature, and 10 J of heat is added to gas A, resulting in a temperature change T. How much heat must be added to gas B to cause the same T? 10 J 100 J 40 J 2.5 J 1600 J For the internal Energy of Ideal Gas For the same number of moles, T depends only on the energy added, not the mass of the gas molecules

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Question 10 The entropy of a closed macroscopic system will never decrease because Energy wouldn’t be conserved if entropy decreased For large systems, the probability of such a decrease is negligible Mechanical equilibrium couldn’t be sustained with a decrease in entropy Heat can never be made to flow from a cold object to a hot object Molecular motions reach their minimum only at absolute 0 The probability of such a macro state occurring is essentially 0

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Question 11 During a certain process, 600 J of heat is added to a system. While this occurs, the system performs 200 J of work on the surroundings. The change in internal energy of the system is: 200 J 800 J 400 J - 400 J Impossible to determine without knowing the temperature The First Law of Thermodynamics

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Question 12 Which of the following is not true about thermal radiation? It’s a mechanism of energy exchange between systems at different temperatures It involves atomic excitations Ice at 00C will emit thermal radiation It requires some material substance to travel through It’s a by product of the food we eat Electromagnetic radiation doesn’t need a material medium to travel through. In E, we eat to maintain body temperature, and we emit a lot of energy as thermal radiation

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Question 13 To increase the diameter of an aluminium ring from 50.0 mm to 50.1 mm, the temperature of the ring must increase by 800C. What temperature change would be necessary to increase the diameter of an aluminium ring from mm to mm? 200C 400C 800C 1100C 1600C We apply the formula for length expansion: Therefore

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Question 14 A gas is enclosed in a metal container with a moveable piston on top. Heat is added to the gas by placing a candle flame in contact with the container's bottom. What of the following is true about the temperature of the gas? The temperature must go up if the piston remains stationary. The temperature must go up if the piston is pulled out dramatically. The temperature must go up no matter what happens to the piston. The temperature must go down no matter what happens to the piston. The temperature must go down if the piston is compressed dramatically. Use the First Law of Thermodynamics, U=Q+W. The temperature adds heat to the gas, so Q is positive. Internal energy is directly related to temperature, so if U is positive, then the temperature goes up (and vice versa). Here we can be sure that U is positive if the work done on the gas is either positive or zero. The only possible answer is A. For B, the work done on the gas is negative because the gas expands (Note: because we add heat to a gas does NOT mean the temperature automatically goes up)

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Question 15 A small heat engine operates using a pan of 1000C boiling water as the high temperature reservoir and the atmosphere as a low temperature reservoir. Assuming ideal behaviour, how much more efficient is the engine on a cold, 00C day than on a warm, 200C day? About 1.3 times as efficient. 2 times as efficient.. 20 times as efficient. Infinitely more efficient. Just as efficient. Using the Engine efficiency equation: Ratio: Hot Day: Cold Day:

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Question 16 In three separate experiments, a gas is transformed from Pi, Vi to state Pf, Vf along paths (1, 2, and 3) illustrated below. The work done on the gas is? Greatest for path 1 Least for path 2 The same for paths 1 and 3 The greatest for path 2 The same for all three paths Vf Vi Pf Pi 2 1 3 D The work done on the gas during a thermodynamic process is equal to the area of the region in the P-V diagram above the V axis and below the path the system makes from its initial to its final state.

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Question 17 A1 m3 container contains 10 moles of ideal gas at room temperature. At what fraction of atmospheric pressure is the gas inside the container? 1/40 atm 1/20 atm 1/10 atm ¼ atm ½ atm V0 2V0 p0 2p0 C A B D Using the Ideal Gas Law Ratio:

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**During an isothermal process, U is ALWAYS ZERO**

Question 18 An ideal gas is compressed isothermally from 20 L to 10 L. During this process, 5J of work done to compress the gas. What is the change of internal energy for this gas? -10 J - 5 J 0 J 5 J 10 J The First Law of Thermodynamics During an isothermal process, U is ALWAYS ZERO

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Question 19 Through a series of thermodynamic processes, the internal energy of a sample of confined gas is increased by 560 J. If the net amount of work done on the sample by its surroundings is 320 J, how much heat was transferred between the gas and its environment? 240 J absorbed 240 J dissipated 880 J absorbed 880 J dissipated None of the above The First Law of Thermodynamics

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Question 20 In one of the steps of the Carnot cycle, the gas undergoes an isothermal expansion. Which of the following statements is true concerning this step? No heat is exchanged between the gas and its surrounding, because the process is isothermal. The temperature decreases because the gas expands. This steps violates the Second Law of Thermodynamics because all the heat absorbed is transferred into work. The internal energy of the gas remains constant. The internal energy of the gas decreases due to the expansion. Statement A is wrong because “no heat exchanged between the gas and its surroundings” is the definition of adiabatic, not isothermal. Statement B cannot be correct since the step described in question is isothermal; by definition, the temperature does not change. Statement C is false, because although the heat absorbed is converted completely to work, it does not include being returned back to it is initial state and having 100% conversion to heat (P decreases and V increases). Isothermal by definition states internal energy remains constant

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Question 21 A Carnot engine using a monatomic ideal gas as a working substance operates between two reservoirs held at 300 K and 200 K, respectively. Starting at point A with pressure and volume as indicated on the graph, the gas expands isothermally to point B, where the volume is 4 x 10-3 m J of heat energy is absorbed by the gas in this process. How many moles of gas are present? Find the work done on the gas during A->B process. Find the work done on the gas during the process B->C adiabatic expansion. How much heat is expelled in the process C->D? Determine the change in internal energy in the process D->A. 2x10-3 4x10-3 3x105 C D B A Isotherm at TH Adiabat: Q=0 Adiabat: Q=0 Isotherm at TC

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Question 21 A Carnot engine using a monatomic ideal gas as a working substance operates between two reservoirs held at 300. K and 200. K, respectively. Starting at point A with pressure and volume as indicated on the graph, the gas expands isothermally to point B, where the volume is 4.0 x 10-3 m J of heat energy is absorbed by the gas in this process. How many moles of gas are present? Find the work done on the gas during A->B process. Find the work done on the gas during the process B->C adiabatic expansion. How much heat is expelled in the process C->D? Determine the change in internal energy in the process D->A. 2.0x10-3 4.0x10-3 3.0x105 C D B A Using the gas law and the given P,V, and T Isotherm at TH Adiabat: Q=0 Adiabat: Q=0 Isotherm at TC

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Question 21 A Carnot engine using a monatomic ideal gas as a working substance operates between two reservoirs held at 300 K and 200 K, respectively. Starting at point A with pressure and volume as indicated on the graph, the gas expands isothermally to point B, where the volume is 4.0 x 10-3 m J of heat energy is absorbed by the gas in this process. How many moles of gas are present? Find the work done on the gas during A->B process. Find the work done on the gas during the process B->C adiabatic expansion. How much heat is expelled in the process C->D? Determine the change in internal energy in the process D->A. 2.0x10-3 4.0x10-3 3.0x105 C D B A Recall: if the temperature is constant, then so is the internal energy. Isotherm at TH Adiabat: Q=0 Along A->B isotherm, there is no change in internal energy, so the first law gives us U = 0 = Q + W. But Q is given as 500 J, so W = -500 J is done ON the gas (the gas DID 500J of work. Adiabat: Q=0 Isotherm at TC

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Question 21 A Carnot engine using a monatomic ideal gas as a working substance operates between two reservoirs held at 300 K and 200 K, respectively. Starting at point A with pressure and volume as indicated on the graph, the gas expands isothermally to point B, where the volume is 4.0 x 10-3 m J of heat energy is absorbed by the gas in this process. How many moles of gas are present? Find the work done on the gas during A->B process. Find the work done on the gas during the process B->C adiabatic expansion. How much heat is expelled in the process C->D? Determine the change in internal energy in the process D->A. Recall: if the process is adiabatic, there is no heat transfer (Q=0). Therefore U=W. 2.0x10-3 4.0x10-3 3.0x105 C D B A Isotherm at TH Now we know the adiabat connects the two isotherms where we know the temperature, and since the ideal gas internal energy depends only upon temperature. Adiabat: Q=0 Adiabat: Q=0 Isotherm at TC W= - 299J

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Question 21 A Carnot engine using a monatomic ideal gas as a working substance operates between two reservoirs held at 300 K and 200 K, respectively. Starting at point A with pressure and volume as indicated on the graph, the gas expands isothermally to point B, where the volume is 4.0 x 10-3 m J of heat energy is absorbed by the gas in this process. How many moles of gas are present? Find the work done on the gas during A->B process. Find the work done on the gas during the process B->C adiabatic expansion. How much heat is expelled in the process C->D? Determine the change in internal energy in the process D->A. We are looking for QC 2.0x10-3 4.0x10-3 3.0x105 C D B A Isotherm at TH For the Carnot cycle: Adiabat: Q=0 Adiabat: Q=0 Also: Isotherm at TC Therefore:

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**The opposite of B->C**

Question 21 A Carnot engine using a monatomic ideal gas as a working substance operates between two reservoirs held at 300 K and 200 K, respectively. Starting at point A with pressure and volume as indicated on the graph, the gas expands isothermally to point B, where the volume is 4.0 x 10-3 m J of heat energy is absorbed by the gas in this process. How many moles of gas are present? Find the work done on the gas during A->B process. Find the work done on the gas during the process B->C adiabatic expansion. How much heat is expelled in the process C->D? Determine the change in internal energy in the process D->A. Once again we are connecting two isotherms. Since the internal energy depends only on temperature, we have 2.0x10-3 4.0x10-3 3.0x105 C D B A Isotherm at TH Adiabat: Q=0 Adiabat: Q=0 Isotherm at TC The opposite of B->C

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Question 22 When a system is taken from state a to state b along the path acb shown in the figure below, 70 J of heat flows into the system, and the system does 30 J of work. When the system is returned from state b to state a along the curved path shown, 60 J of heat flows out of the system. Does the system perform work on its surroundings or do the surroundings perform work on the system? How much work is done? If the system does 10 J of work in transforming from state a to state b along path adb, does the system absorb or does it emit heat? How much heat is transferred? If Ua=0 J and Ud=30 J, determine the heat absorbed in the process db and ad. For the process adbca, identify each of the following quantities as positive, negative, or zero. W = _________________ Q = _________________ U = _________________ V P b c d a

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**Question 22 b c P We require the value of Wba**

When a system is taken from state a to state b along the path acb shown in the figure below, 70 J of heat flows into the system, and the system does 30 J of work. When the system is returned from state b to state a along the curved path shown, 60 J of heat flows out of the system. Does the system perform work on its surroundings or do the surroundings perform work on the system? How much work is done? V P b c d a We require the value of Wba First let’s calculate Uabc -30 J because system does work Because Ua->b does not depend on path taken from a to b, then since Uab = 40J then Uba = -Uab = – 40 J Since the value is + 20J, the surroundings does the work on the gas (system).

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Question 22 When a system is taken from state a to state b along the path acb shown in the figure below, 70 J of heat flows into the system, and the system does 30 J of work. If the system does 10 J of work in transforming from state a to state b along path adb, does the system absorb or does it emit heat? How much heat is transferred? V P b c d a Again, using the fact that Ua->b does not depend on the path taken, we know that Uadb = 40 J Now we require Qadb Because the system does 10 J of work Therefore the system absorbs 50 J of heat

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Question 22 When a system is taken from state a to state b along the path acb shown in the figure below, 70 J of heat flows into the system, and the system does 30 J of work. C) If Ua=0 J and Ud=30 J, determine the heat absorbed in the process db and ad. V P b c d a For the process db, there is no change in volume, therefore Wdb = 0J. Now, Uab = 40 J, and Ua = 0 J, tells us that Ub = 40 J, therefore: For ad: Wadb=Wad+Wdb=Wad+0J=Wad Since Wadb =-10 J, then Wad =-10J

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**Question 22 b c P d a The process adbca is cyclic, so U = zero V**

When a system is taken from state a to state b along the path acb shown in the figure below, 70 J of heat flows into the system, and the system does 30 J of work. For the process adbca, identify each of the following quantities as positive, negative, or zero. W = _________________ Q = _________________ U = _________________ V P b c d a The process adbca is cyclic, so U = zero Because the process is traversed counter-clockwise in the PV diagram , we know that W is positive (on the system). The system is a heat engine, emitting heat on each cycle U=Q+W therefore Q must be negative

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Question 23 1 mole of monatomic ideal gas is brought through a cycle A->B->C->D->A as depicted below. All processes are performed slowly. Respond to the following in terms of P0, V0, and R. Find the temperature at each vertex. Find the heat added to the gas for the process A->B. Find the work done on the gas for the process C->D. Find the heat added to the gas fro the process D->A Find the change in internal energy for Process B->C The entire cycle V0 2V0 P0 2P0 C A B D 3V0

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**Question 23 V0 2V0 P0 2P0 C A B D 3V0 From the Ideal Gas Law:**

1 mole of monatomic ideal gas is brought through a cycle A->B->C->D->A as depicted below. All processes are performed slowly. Respond to the following in terms of P0, V0, and R. Find the temperature at each vertex. Find the heat added to the gas for the process A->B. Find the work done on the gas for the process C->D. Find the heat added to the gas fro the process D->A Find the change in internal energy for Process B->C The entire cycle V0 2V0 P0 2P0 C A B D 3V0 From the Ideal Gas Law:

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Question 23 1 mole of monatomic ideal gas is brought through a cycle A->B->C->D->A as depicted below. All processes are performed slowly. Respond to the following in terms of P0, V0, and R. Find the temperature at each vertex. Find the heat added to the gas for the process A->B. Find the work done on the gas for the process C->D. Find the heat added to the gas fro the process D->A Find the change in internal energy for Process B->C The entire cycle V0 2V0 P0 2P0 C A B D 3V0 Process A->B occurs at constant pressure. So we can use the First Law Since Work done on the gas is negative of the area under graph

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Question 23 1 mole of monatomic ideal gas is brought through a cycle A->B->C->D->A as depicted below. All processes are performed slowly. Respond to the following in terms of P0, V0, and R. Find the temperature at each vertex. Find the heat added to the gas for the process A->B. Find the work done on the gas for the process C->D. Find the heat added to the gas fro the process D->A Find the change in internal energy for Process B->C The entire cycle V0 2V0 P0 2P0 C A B D 3V0 The area under the process C->D The sign of the net work is negative for cycles that run clockwise The decrease in volume corresponds to positive work done on the gas

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Question 23 1 mole of monatomic ideal gas is brought through a cycle A->B->C->D->A as depicted below. All processes are performed slowly. Respond to the following in terms of P0, V0, and R. Find the temperature at each vertex. Find the heat added to the gas for the process A->B. Find the work done on the gas for the process C->D. Find the heat added to the gas fro the process D->A Find the change in internal energy for Process B->C The entire cycle V0 2V0 P0 2P0 C A B D 3V0 Process D->A occurs at constant volume. So we can use the First Law Since at constant volume, no work is done

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**Question 23 V0 2V0 P0 2P0 C A B D 3V0 i) Using:**

1 mole of monatomic ideal gas is brought through a cycle A->B->C->D->A as depicted below. All processes are performed slowly. Respond to the following in terms of P0, V0, and R. Find the temperature at each vertex. Find the heat added to the gas for the process A->B. Find the work done on the gas for the process C->D. Find the heat added to the gas fro the process D->A. Find the change in internal energy for Process B->C The entire cycle V0 2V0 P0 2P0 C A B D 3V0 i) Using: ii) For the entire cycle, you end up back in the same state, so U = 0

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**Question 24 b a c V (x10-3 m3) P (x105 Pa) 2.4 0.6 12 48**

A 0.4 mol sample of an ideal diatomic gas undergoes slow changes from state a to state b to state c and back to a along the cycle shown. Path ab is an isotherm and the work done by the gas as it changes isothermally from state a to state b is given by the equation: What is the temperature of: State a State b State c Determine the change in internal energy of the gas for: Step ab. Step bc. Step ca. How much work, Wab is done by the gas during step ab? What is the total work done over the cycle abca? Is heat absorbed or discarded step ab? If so how much? What is the maximum possible efficiency (without violating the Second Law of Thermodynamics) for a cyclical heat engine that operates between the temperatures of state a and c? b a c V (x10-3 m3) P (x105 Pa) 2.4 0.6 12 48

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**Question 24 b a c V (x10-3 m3) P (x105 Pa) 2.4 0.6 12 48**

A 0.4 mol sample of an ideal diatomic gas undergoes slow changes from state a to state b to state c and back to a along the cycle shown. Path ab is an isotherm and the work done by the gas as it changes isothermally from state a to state b is given by the equation: What is the temperature of: State a State b State c b a c V (x10-3 m3) P (x105 Pa) 2.4 0.6 12 48 From the Ideal Gas Law: State b is on the Isotherm with state a

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**Question 24 b a c V (x10-3 m3) P (x105 Pa) 2.4 0.6 12 48**

A 0.4 mol sample of an ideal diatomic gas undergoes slow changes from state a to state b to state c and back to a along the cycle shown. Path ab is an isotherm and the work done by the gas as it changes isothermally from state a to state b is given by the equation: Determine the change in internal energy of the gas for: Step ab. Step bc. Step ca. b a c V (x10-3 m3) P (x105 Pa) 2.4 0.6 12 48 Since step a takes place along an isotherm, Uab = 0 or or

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**Question 24 b a c V (x10-3 m3) P (x105 Pa) 2.4 0.6 12 48**

A 0.4 mol sample of an ideal diatomic gas undergoes slow changes from state a to state b to state c and back to a along the cycle shown. Path ab is an isotherm and the work done by the gas as it changes isothermally from state a to state b is given by the equation: C) How much work, Wab is done by the gas during step ab? b a c V (x10-3 m3) P (x105 Pa) 2.4 0.6 12 48

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**Question 24 b a c V (x10-3 m3) P (x105 Pa) 2.4 0.6 12 48**

A 0.4 mol sample of an ideal diatomic gas undergoes slow changes from state a to state b to state c and back to a along the cycle shown. Path ab is an isotherm and the work done by the gas as it changes isothermally from state a to state b is given by the equation: D) What is the total work done over the cycle abca? b a c V (x10-3 m3) P (x105 Pa) 2.4 0.6 12 48 The total work done over a cycle is equal to the sum of the values of the work done over each step.

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**Question 24 b a c V (x10-3 m3) P (x105 Pa) 2.4 0.6 12 48**

A 0.4 mol sample of an ideal diatomic gas undergoes slow changes from state a to state b to state c and back to a along the cycle shown. Path ab is an isotherm and the work done by the gas as it changes isothermally from state a to state b is given by the equation: Is heat absorbed or discarded step ab? If so how much? b a c V (x10-3 m3) P (x105 Pa) 2.4 0.6 12 48 By the First Law of Thermodynamics Since Q is positive, this represents heat absorbed by the gas.

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**Question 24 b a c V (x10-3 m3) P (x105 Pa) 2.4 0.6 12 48**

A 0.4 mol sample of an ideal diatomic gas undergoes slow changes from state a to state b to state c and back to a along the cycle shown. Path ab is an isotherm and the work done by the gas as it changes isothermally from state a to state b is given by the equation: G) What is the maximum possible efficiency (without violating the Second Law of Thermodynamics) for a cyclical heat engine that operates between the temperatures of state a and c? b a c V (x10-3 m3) P (x105 Pa) 2.4 0.6 12 48 The maximum possible efficiency is the efficiency of the Carnot engine.

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Question 25 You have just purchased a cup of coffee (45 g), but you notice that the coffee is cold, so you call over your host. “It’s 45oC”, you say, “But I like it 65oC. “. The coffee in the pot is 95oC, how much coffee should the host pour to raise the temperature of the 45g of coffee in your cup? Look at the heat lost by the pot’s coffee and the heat gained by your coffee Coffee in cup: Now, Coffee in pot:

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