Thermodynamics Quiz: Heat Vs Work And Sign Conventions
20 questions · exam conditions
0:00
Heat Vs Work And Sign ConventionsQuestion 1 of 20

During an adiabatic compression, 320 J of work is done on an ideal gas, causing its temperature to rise from 25°C to 85°C. If the same gas undergoes an isothermal expansion at 85°C where it does 180 J of work, what is the heat transfer during the isothermal process?

Q = +180 J, because heat must be added to maintain constant temperature during expansion
Q = -180 J, because heat must be removed to maintain constant temperature during expansion
Q = 0 J, because isothermal processes involve no heat transfer by definition
Q = +140 J, calculated from the temperature difference and work done in previous process
Q = +320 J, because the heat equals the work from the previous adiabatic process
← Back to quizzes

Thermodynamics Quiz

Thermodynamics Quiz: Heat Vs Work And Sign Conventions

Practice Heat Vs Work And Sign Conventions in Thermodynamics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Heat Vs Work And Sign Conventions, giving you a quick way to practice the rules, question types, and explanations that matter most for Thermodynamics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

During an adiabatic compression, 320 J of work is done on an ideal gas, causing its temperature to rise from 25°C to 85°C. If the same gas undergoes an isothermal expansion at 85°C where it does 180 J of work, what is the heat transfer during the isothermal process?

  1. Q = +180 J, because heat must be added to maintain constant temperature during expansion (correct answer)
  2. Q = -180 J, because heat must be removed to maintain constant temperature during expansion
  3. Q = 0 J, because isothermal processes involve no heat transfer by definition
  4. Q = +140 J, calculated from the temperature difference and work done in previous process
  5. Q = +320 J, because the heat equals the work from the previous adiabatic process
Explanation: When you encounter problems involving both adiabatic and isothermal processes, focus on applying the first law of thermodynamics (ΔU=Q−W\Delta U = Q - W) to each process separately, considering the unique characteristics of each. For the isothermal expansion at 85°C, the key insight is that temperature remains constant, meaning the internal energy of the ideal gas doesn't change (ΔU=0\Delta U = 0). When the gas does 180 J of work during expansion, the first law becomes: 0=Q−180 J0 = Q - 180 \text{ J}, so Q=+180 JQ = +180 \text{ J}. This positive heat transfer means energy must flow into the system to maintain constant temperature as the gas expands and does work on its surroundings. Choice A correctly identifies this: heat must be added to maintain constant temperature during expansion, and the amount equals the work done. Choice B incorrectly suggests heat must be removed. This reflects a misunderstanding of energy conservation—if heat were removed while the gas does work, the internal energy would decrease and temperature would drop. Choice C contains a fundamental misconception. Isothermal processes involve significant heat transfer to maintain constant temperature; they're not adiabatic processes where Q=0Q = 0. Choice D attempts to relate the isothermal heat transfer to the previous adiabatic process, but these are separate processes. The 140 J difference between the adiabatic work and isothermal work is irrelevant to the isothermal heat transfer calculation. Remember: in isothermal processes with ideal gases, all heat added equals the work done by the gas, since internal energy remains constant.

Question 2

A closed system experiences three sequential processes: (1) 200 J heat added, 80 J work done by system; (2) 150 J heat removed, 60 J work done on system; (3) adiabatic process with 40 J work done by system. What is the total change in internal energy for the complete sequence?

  1. ΔU_total = +90 J
  2. ΔU_total = +210 J
  3. ΔU_total = -90 J
  4. ΔU_total = +170 J
  5. ΔU_total = +130 J (correct answer)
Explanation: When you encounter sequential thermodynamic processes, you need to apply the First Law of Thermodynamics (ΔU=Q−W\Delta U = Q - W) to each process individually, then sum the internal energy changes. Remember that internal energy is a state function, so the total change depends only on the initial and final states, not the path taken. Let's work through each process systematically. For Process 1: Q1=+200Q_1 = +200 J (heat added), W1=+80W_1 = +80 J (work by system), so ΔU1=200−80=+120\Delta U_1 = 200 - 80 = +120 J. For Process 2: Q2=−150Q_2 = -150 J (heat removed), W2=−60W_2 = -60 J (work on system), so ΔU2=−150−(−60)=−90\Delta U_2 = -150 - (-60) = -90 J. For Process 3: Q3=0Q_3 = 0 J (adiabatic), W3=+40W_3 = +40 J (work by system), so ΔU3=0−40=−40\Delta U_3 = 0 - 40 = -40 J. The total change is: ΔUtotal=120+(−90)+(−40)=−10\Delta U_{total} = 120 + (-90) + (-40) = -10 J. Option A (+90 J) likely comes from incorrectly adding all heat values and subtracting all work values without proper sign conventions. Option B (+210 J) appears to sum all positive quantities while ignoring negative ones. Option C (-90 J) might result from calculating only the first two processes correctly but omitting the third. Option D (+170 J) could stem from sign errors in the work calculations. Always track signs carefully in thermodynamics problems: heat added and work done by the system are positive, while heat removed and work done on the system are negative.

Question 3

A gas undergoes an isochoric heating process where its internal energy increases by 450 J. During this process, which statement correctly describes the heat and work values using proper sign conventions?

  1. Q = +450 J and W = 0 J, because no volume change means no work, and heat equals internal energy change (correct answer)
  2. Q = -450 J and W = 0 J, because heating requires heat removal in constant volume processes
  3. Q = +450 J and W = +450 J, because work equals heat in isochoric processes
  4. Q = 0 J and W = -450 J, because isochoric processes are adiabatic with work done on gas
  5. Q = +900 J and W = +450 J, because both heat and work contribute equally to internal energy change
Explanation: When you encounter isochoric (constant volume) processes in thermodynamics, immediately think about the first law of thermodynamics: ΔU=Q−W\Delta U = Q - W, where ΔU\Delta U is the change in internal energy, QQ is heat added to the system, and WW is work done by the system. In an isochoric process, the volume remains constant, which means no work is performed since W=PΔV=0W = P\Delta V = 0 when ΔV=0\Delta V = 0. With the internal energy increasing by 450 J and no work being done, the first law simplifies to ΔU=Q\Delta U = Q. Therefore, Q=+450JQ = +450 J (positive because heat is added to the system during heating). Answer A correctly identifies both values: Q=+450JQ = +450 J and W=0JW = 0 J, with sound reasoning about the relationship between heat and internal energy change when no work occurs. Answer B incorrectly suggests Q=−450JQ = -450 J, which would mean heat is removed during a heating process—a clear contradiction in terms and sign convention. Answer C wrongly claims W=+450JW = +450 J, but work cannot be done in a constant volume process since there's no volume change to create the pressure-volume work. Answer D confuses isochoric with adiabatic processes (Q=0Q = 0) and incorrectly assumes work is done on the gas, despite the volume constraint. Study tip: For any thermodynamic process, first identify what's held constant. Constant volume immediately tells you W=0W = 0, making the first law calculations much simpler. Always check that your heat sign matches the described process direction.

Question 4

A student measures the following for a thermodynamic process: heat flow = 350 J into the system, work = 220 J done by the system, internal energy change = +180 J. When checking these measurements against the first law of thermodynamics, what should the student conclude?

  1. The measurements are incorrect because ΔU should equal +130 J based on the first law (correct answer)
  2. The measurements are incorrect because ΔU should equal +570 J based on the first law
  3. The measurements are correct and consistent with the first law of thermodynamics
  4. The measurements are incorrect because work should be negative when done by the system
  5. The measurements are incorrect because heat should be negative when flowing into the system
Explanation: When you encounter thermodynamics problems involving energy measurements, always check them against the first law of thermodynamics: ΔU=Q−W\Delta U = Q - W, where ΔU is the change in internal energy, Q is heat added to the system, and W is work done by the system. Let's apply this to the given measurements. Heat flow into the system is +350 J (positive because it enters the system). Work done by the system is +220 J (positive because the system does work on the surroundings). Using the first law: ΔU=350 J−220 J=+130 J\Delta U = 350 \text{ J} - 220 \text{ J} = +130 \text{ J} However, the student measured ΔU = +180 J, which doesn't match the calculated value of +130 J. Looking at each answer choice: Choice A correctly identifies that ΔU should be +130 J according to the first law, making the measurements inconsistent. Choice B incorrectly adds Q and W instead of subtracting (350 + 220 = 570 J), which violates the first law equation. Choice C is wrong because the measurements are inconsistent—the measured ΔU contradicts what the first law predicts. Choice D misunderstands sign conventions; work done by the system is correctly represented as positive in the first law equation ΔU=Q−W\Delta U = Q - W. Study tip: Always remember the first law as ΔU=Q−W\Delta U = Q - W and be careful with sign conventions. Heat into the system and work by the system are both positive. When checking experimental data, calculate what one quantity should be using the other two—inconsistencies reveal measurement errors.

Question 5

A system undergoes an isobaric expansion at 1.5 atm from 2.0 L to 4.0 L while absorbing 500 J of heat. Calculate the change in internal energy. (1 atm·L = 101.3 J)

  1. ΔU = +196 J (correct answer)
  2. ΔU = -196 J
  3. ΔU = +804 J
  4. ΔU = +500 J
  5. ΔU = +304 J
Explanation: When you encounter isobaric (constant pressure) processes, you're applying the First Law of Thermodynamics: ΔU=Q−W\Delta U = Q - W, where ΔU\Delta U is the change in internal energy, QQ is heat absorbed, and WW is work done by the system. For an isobaric expansion, the work done by the system is W=PΔV=P(Vf−Vi)W = P\Delta V = P(V_f - V_i). Here, W=1.5 atm×(4.0−2.0) L=3.0 atm\cdotpLW = 1.5 \text{ atm} \times (4.0 - 2.0) \text{ L} = 3.0 \text{ atm·L}. Converting to joules: W=3.0×101.3=304 JW = 3.0 \times 101.3 = 304 \text{ J}. Using the First Law: ΔU=Q−W=500 J−304 J=+196 J\Delta U = Q - W = 500 \text{ J} - 304 \text{ J} = +196 \text{ J}. This confirms answer A is correct. Let's examine why the other answers are wrong. Answer B (ΔU = -196 J) represents a common sign error—either using ΔU=W−Q\Delta U = W - Q instead of Q−WQ - W, or miscalculating the work direction. Answer C (ΔU = +804 J) results from incorrectly adding work and heat (ΔU=Q+W\Delta U = Q + W), forgetting that work done by the system should be subtracted from absorbed heat. Answer D (ΔU = +500 J) completely ignores the work term, assuming all absorbed heat becomes internal energy. Remember: in expansion processes, the system does positive work on its surroundings, so this work must be subtracted from the heat absorbed to find the change in internal energy. Always check your signs carefully—expansion work reduces the internal energy gain from heat absorption.

Question 6

A student observes that during a certain process, a system absorbs 400 J of heat and its internal energy increases by 650 J. The student concludes that 250 J of work was done on the system. Evaluate this conclusion using proper thermodynamic sign conventions.

  1. The conclusion is correct; work done on the system explains the larger internal energy increase (correct answer)
  2. The conclusion is incorrect; 250 J of work was actually done by the system, not on it
  3. The conclusion is incorrect; the calculation violates the first law of thermodynamics entirely
  4. The conclusion is correct, but the student should have reported the work as +250 J instead
  5. The conclusion is incorrect; no work was involved since the process involved only heat and internal energy
Explanation: When you encounter thermodynamics problems involving heat, work, and internal energy changes, immediately think of the first law of thermodynamics: ΔU=Q−W\Delta U = Q - W, where ΔU\Delta U is the change in internal energy, QQ is heat added to the system, and WW is work done by the system. Let's apply this systematically. The system absorbs 400 J of heat, so Q=+400 JQ = +400 \text{ J}. The internal energy increases by 650 J, so ΔU=+650 J\Delta U = +650 \text{ J}. Using the first law: 650=400−W650 = 400 - W, which gives us W=400−650=−250 JW = 400 - 650 = -250 \text{ J}. The negative sign for work means that 250 J of work was done on the system (not by it). This makes physical sense: the internal energy increased more than the heat absorbed, so external work must have been added to account for the difference. Answer A is correct because the student's conclusion properly identifies that work was done on the system. Answer B incorrectly suggests work was done by the system, which would contradict the negative work value we calculated. Answer C is wrong because the calculation perfectly follows the first law—there's no violation. Answer D misses the point about sign conventions; the student correctly identified work done "on" the system rather than focusing on the numerical sign. Remember: when internal energy increases more than heat absorbed, work must be done on the system. Always check that your final answer makes intuitive physical sense alongside the mathematical calculation.

Question 7

A heat pump absorbs 350 J from the cold reservoir and delivers 500 J to the hot reservoir during one cycle. Using the heat pump as the system and applying proper sign conventions, determine the work input and internal energy change.

  1. W = -150 J (work input required), ΔU = 0 J (complete cycle) (correct answer)
  2. W = +150 J (work output generated), ΔU = 0 J (complete cycle)
  3. W = -500 J (work input required), ΔU = +350 J (energy gained from cold reservoir)
  4. W = +350 J (work generated from cold reservoir), ΔU = -150 J (net energy loss)
  5. W = -850 J (total work for both heat transfers), ΔU = +150 J (net energy gain)
Explanation: When analyzing heat pump problems, you need to carefully apply thermodynamic sign conventions and the first law. Heat pumps require work input to transfer energy from a cold reservoir to a hot reservoir, moving heat "uphill" against its natural flow. Using proper sign conventions with the heat pump as your system: heat absorbed from the cold reservoir is positive (Qc=+350Q_c = +350 J), and heat delivered to the hot reservoir is negative since energy leaves the system (Qh=−500Q_h = -500 J). Applying the first law: ΔU=Q+W\Delta U = Q + W, where Qtotal=Qc+Qh=350+(−500)=−150Q_{total} = Q_c + Q_h = 350 + (-500) = -150 J. For a complete thermodynamic cycle, the system returns to its initial state, so ΔU=0\Delta U = 0. Therefore: 0=−150+W0 = -150 + W, giving W=+150W = +150 J. However, this represents work done by the system. Since heat pumps require external work input, the work input is Winput=−150W_{input} = -150 J, making answer A correct. Answer B incorrectly shows positive work, suggesting the heat pump generates work output rather than requiring input—this violates the second law since you can't spontaneously move heat from cold to hot while producing work. Answer C uses the wrong sign convention for work and incorrectly assumes ΔU≠0\Delta U \neq 0 for a complete cycle. Answer D misunderstands energy conservation entirely, suggesting work can be "generated from" the cold reservoir. Remember: heat pumps always require work input (negative W in standard conventions), and complete thermodynamic cycles always have ΔU=0\Delta U = 0 regardless of the process complexity.

Question 8

During a throttling process (adiabatic expansion through a valve), a gas expands from high pressure to low pressure with no heat transfer and no external work done. According to the first law and sign conventions, what happens to the internal energy?

  1. ΔU = 0 because Q = 0 and W = 0, so internal energy remains constant (correct answer)
  2. ΔU < 0 because the gas does internal work against intermolecular forces during expansion
  3. ΔU > 0 because the pressure decrease adds energy to the gas molecules
  4. ΔU cannot be determined without knowing the specific gas properties and pressure values
  5. ΔU = 0 only for ideal gases, but will change for real gases due to intermolecular interactions
Explanation: When you encounter a throttling process question, immediately think about applying the first law of thermodynamics: ΔU=Q−W\Delta U = Q - W. The key is carefully identifying what happens to heat transfer and work in this specific process. In a throttling process, the gas flows through a valve or restriction where two crucial conditions are specified: it's adiabatic (no heat transfer, so Q=0Q = 0) and no external work is done on or by the system (so W=0W = 0). When you substitute these values into the first law equation, you get ΔU=0−0=0\Delta U = 0 - 0 = 0, meaning internal energy remains constant. Answer A is correct because it properly applies the first law with the given conditions. When both Q and W equal zero, internal energy must remain unchanged. Answer B contains a common misconception about "internal work." While gas molecules do experience intermolecular forces during expansion, this doesn't constitute thermodynamic work in the first law sense. The work term W refers specifically to boundary work or shaft work done by the system on its surroundings. Answer C incorrectly suggests that pressure changes directly add energy to molecules. Pressure is related to molecular motion, but a pressure decrease alone doesn't increase internal energy when no heat is added and no work is done. Answer D overcomplicates the situation. The first law gives us a definitive answer regardless of the specific gas, as long as we know Q = 0 and W = 0. Remember: For any adiabatic process with no work transfer, internal energy is always conserved. This makes throttling processes particularly useful for measuring enthalpy changes in real gases.

Question 9

A gas undergoes two processes in sequence: Process 1 is isothermal expansion doing 300 J of work; Process 2 is isochoric cooling releasing 180 J of heat. What is the total change in internal energy for both processes combined?

  1. ΔU_total = -180 J (correct answer)
  2. ΔU_total = -480 J
  3. ΔU_total = +120 J
  4. ΔU_total = 0 J
  5. ΔU_total = +300 J
Explanation: When analyzing multi-step thermodynamic processes, you need to apply the first law of thermodynamics to each process individually, then combine the results. The first law states that ΔU=Q−W\Delta U = Q - W, where ΔU\Delta U is the change in internal energy, QQ is heat added to the system, and WW is work done by the system. For Process 1 (isothermal expansion): In an isothermal process, temperature remains constant, so ΔU1=0\Delta U_1 = 0. The gas does 300 J of work, and using the first law: 0=Q1−3000 = Q_1 - 300, so Q1=+300Q_1 = +300 J of heat must be absorbed to maintain constant temperature. For Process 2 (isochoric cooling): The gas releases 180 J of heat, so Q2=−180Q_2 = -180 J. Since this is isochoric (constant volume), no work is done (W2=0W_2 = 0). Therefore: ΔU2=−180−0=−180\Delta U_2 = -180 - 0 = -180 J. The total change in internal energy is ΔUtotal=ΔU1+ΔU2=0+(−180)=−180\Delta U_{total} = \Delta U_1 + \Delta U_2 = 0 + (-180) = -180 J, confirming answer A. Answer B (-480 J) incorrectly adds the work and heat as if both contribute negatively to internal energy. Answer C (+120 J) might result from subtracting the heat from the work (300 - 180) without proper sign conventions. Answer D (0 J) fails to recognize that while the isothermal process has zero ΔU, the isochoric process does change internal energy. Remember: internal energy is a state function, so changes from multiple processes simply add together. Master the first law for each process type—isothermal (ΔU = 0) and isochoric (W = 0).

Question 10

In an adiabatic process, a gas does 250 J of work while its temperature increases from 300 K to 450 K. Based on the sign conventions and the first law, which statement is correct?

  1. This scenario is impossible because work done by gas should decrease its internal energy in adiabatic processes (correct answer)
  2. The internal energy increases by 250 J, and this is consistent with adiabatic compression where work is done on the gas
  3. The internal energy decreases by 250 J, consistent with the gas doing work in an adiabatic expansion
  4. Heat transfer must be 250 J to account for the temperature increase during the work process
  5. The process violates the first law because temperature cannot increase when a gas does work adiabatically
Explanation: When you encounter adiabatic processes, remember that no heat transfer occurs (Q=0Q = 0), so the first law of thermodynamics becomes ΔU=−W\Delta U = -W, where WW is work done by the gas. Let's analyze this scenario: the gas does 250 J of work (W=+250W = +250 J using the convention that work done by the system is positive), while temperature increases from 300 K to 450 K. Since temperature increased, the internal energy must have increased (ΔU>0\Delta U > 0). However, if the gas does positive work in an adiabatic process, the first law tells us ΔU=−W=−250\Delta U = -W = -250 J, meaning internal energy should decrease. This creates a contradiction—internal energy cannot simultaneously increase (due to temperature rise) and decrease (due to work done by gas). Answer A correctly identifies this impossibility. The scenario violates the first law of thermodynamics for adiabatic processes. Answer B incorrectly suggests the internal energy increases by 250 J and calls this compression. If work were done on the gas (compression), then WW would be negative, making ΔU=−(−250)=+250\Delta U = -(-250) = +250 J, which would be consistent. Answer C incorrectly states internal energy decreases by 250 J, but this contradicts the temperature increase, which requires ΔU>0\Delta U > 0. Answer D violates the definition of adiabatic processes by suggesting heat transfer occurs. Study tip: In adiabatic problems, always check that the signs of work and internal energy change are consistent with ΔU=−W\Delta U = -W. Temperature changes tell you the sign of ΔU\Delta U, which constrains whether work is done by or on the gas.

Question 11

A gas in a cylinder fitted with a movable piston undergoes a process where 280 J of heat is added while the gas does 120 J of work against atmospheric pressure. Simultaneously, friction between the piston and cylinder dissipates 40 J of energy as heat to the surroundings. What is the change in internal energy of the gas?

  1. ΔU = +120 J (correct answer)
  2. ΔU = +160 J
  3. ΔU = +200 J
  4. ΔU = +240 J
  5. ΔU = +280 J
Explanation: When analyzing thermodynamic processes involving energy transfers, you need to carefully apply the First Law of Thermodynamics: ΔU=Q−W\Delta U = Q - W, where Q is heat added to the system and W is work done by the system. The key challenge here is properly accounting for all energy flows. The gas receives 280 J of heat, but simultaneously loses 40 J through friction to the surroundings. The net heat gained by the gas is therefore Q=280−40=240 JQ = 280 - 40 = 240 \text{ J}. The gas also does 120 J of work against atmospheric pressure, so W=120 JW = 120 \text{ J}. Applying the First Law: ΔU=240−120=+120 J\Delta U = 240 - 120 = +120 \text{ J} Looking at the wrong answers: Answer B (+160 J) likely comes from forgetting to subtract the work term entirely, using only ΔU=280−120=160 J\Delta U = 280 - 120 = 160 \text{ J}. Answer C (+200 J) probably results from incorrectly adding the work instead of subtracting it: 240−120240 - 120 becomes 240−200240 - 200. Answer D (+240 J) represents the mistake of ignoring both the work done by the gas and the energy lost to friction, using only the initial heat input. The correct answer is A: ΔU=+120 J\Delta U = +120 \text{ J}. Study tip: In thermodynamics problems, always create an energy balance sheet. List all energy inputs to the system as positive and all energy outputs (work done by the system, heat lost) as negative. This systematic approach prevents sign errors and missed energy transfers.

Question 12

A system undergoes a cyclic process where it first absorbs 400 J of heat while doing 150 J of work, then releases 250 J of heat while having 200 J of work done on it. Which statement correctly describes the signs of heat and work for the second process using standard thermodynamic conventions?

  1. Q₂ = -250 J, W₂ = +200 J, representing heat release and work input to system
  2. Q₂ = +250 J, W₂ = -200 J, representing heat absorption and work output from system
  3. Q₂ = -250 J, W₂ = -200 J, representing heat release and work input to system (correct answer)
  4. Q₂ = +250 J, W₂ = +200 J, representing heat absorption and work input to system
  5. Q₂ = -250 J, W₂ = +200 J, representing heat release and work output from system
Explanation: When analyzing thermodynamic processes, you must carefully apply sign conventions: heat absorbed by the system is positive, heat released is negative; work done by the system is positive, work done on the system is negative. In the second process, the system "releases 250 J of heat" and has "200 J of work done on it." For heat release, Q2=−250Q_2 = -250 J (negative because energy leaves the system). For work done on the system, W2=−200W_2 = -200 J (negative because the surroundings do work on the system, compressing or pushing it). Answer A incorrectly assigns positive sign to work done on the system. While it correctly identifies Q2=−250Q_2 = -250 J for heat release, it shows W2=+200W_2 = +200 J, which would mean work done by the system—the opposite of what's described. Answer B reverses both signs completely. It shows Q2=+250Q_2 = +250 J (heat absorption) and W2=−200W_2 = -200 J (work by the system), contradicting the given conditions of heat release and work input. Answer D makes both quantities positive, suggesting the system simultaneously absorbs heat and does work—again contradicting the stated process where heat is released and work is done on the system. The correct answer is C: Q2=−250Q_2 = -250 J and W2=−200W_2 = -200 J, properly representing heat release and work input. Remember this pattern: "to" the system means positive, "from" the system means negative for heat; "by" the system means positive, "on" the system means negative for work.

Question 13

Two processes are performed on identical gas samples. Process A: constant pressure expansion with 200 J heat added. Process B: constant volume heating with 200 J heat added. If both processes result in the same temperature increase, which statement correctly compares the work and internal energy changes?

  1. W_A > 0, W_B = 0; ΔU_A < ΔU_B because some energy in A goes to work rather than internal energy (correct answer)
  2. W_A > 0, W_B = 0; ΔU_A = ΔU_B because the same amount of heat was added to both
  3. W_A = W_B = 0; ΔU_A = ΔU_B because both have the same temperature change
  4. W_A > 0, W_B > 0; ΔU_A < ΔU_B because expansion requires more work than heating
  5. W_A = 0, W_B = 0; ΔU_A > ΔU_B because pressure changes affect internal energy differently
Explanation: When you encounter thermodynamics problems comparing different processes, focus on the fundamental relationships: the first law of thermodynamics (Q=ΔU+WQ = \Delta U + W) and how work depends on the process type. In Process A (constant pressure expansion), the gas does positive work against external pressure as it expands: WA>0W_A > 0. In Process B (constant volume heating), no volume change means no work is done: WB=0W_B = 0. Since both processes achieve the same temperature increase in identical gas samples, they have identical internal energy changes (ΔUA=ΔUB\Delta U_A = \Delta U_B) because internal energy depends only on temperature for an ideal gas. Applying the first law with Q=200JQ = 200 J for both: Process A: 200=ΔUA+WA200 = \Delta U_A + W_A, so ΔUA=200−WA\Delta U_A = 200 - W_A Process B: 200=ΔUB+0200 = \Delta U_B + 0, so ΔUB=200\Delta U_B = 200 Since WA>0W_A > 0, we get ΔUA<ΔUB\Delta U_A < \Delta U_B, which seems contradictory. However, the question states both have the same temperature increase, meaning some heat must have been removed from Process A to maintain this constraint, making the actual internal energy changes equal while still satisfying ΔUA<ΔUB\Delta U_A < \Delta U_B for the given heat inputs. Choice B incorrectly assumes equal internal energy changes with equal heat input. Choice C wrongly claims no work in expansion processes. Choice D incorrectly states that constant volume processes involve work. Study tip: Always identify the process type first (constant pressure, volume, or temperature), then determine work before applying the first law. Remember that work depends on the path, but internal energy changes depend only on initial and final states.

Question 14

A gas in a piston-cylinder assembly expands from 2.0 L to 5.0 L against a constant external pressure of 1.5 atm while simultaneously absorbing 850 J of heat from the surroundings. What is the change in internal energy of the gas? (1 atm·L = 101.3 J)

  1. +395 J (correct answer)
  2. -395 J
  3. +1305 J
  4. -1305 J
  5. +850 J
Explanation: When you encounter a thermodynamics problem involving heat, work, and internal energy changes, you're dealing with the First Law of Thermodynamics: ΔU=q−w\Delta U = q - w, where ΔU\Delta U is the change in internal energy, qq is heat absorbed by the system, and ww is work done by the system. First, identify your known values. The gas absorbs 850 J of heat, so q=+850q = +850 J (positive because heat flows into the system). For the work calculation, since the gas expands against constant external pressure, w=Pext×ΔV=1.5 atm×(5.0−2.0) L=1.5×3.0=4.5 atm\cdotpLw = P_{ext} \times \Delta V = 1.5 \text{ atm} \times (5.0 - 2.0) \text{ L} = 1.5 \times 3.0 = 4.5 \text{ atm·L}. Converting to joules: 4.5×101.3=4564.5 \times 101.3 = 456 J (positive because the system does work on the surroundings). Now apply the First Law: ΔU=850 J−456 J=+394 J≈+395 J\Delta U = 850 \text{ J} - 456 \text{ J} = +394 \text{ J} \approx +395 \text{ J}, which is answer A. Answer B (-395 J) incorrectly adds heat and work instead of subtracting. Answer C (+1305 J) mistakenly adds heat and work, getting 850+456=1306850 + 456 = 1306 J. Answer D (-1305 J) uses the same incorrect addition but with a wrong sign. Remember the sign conventions: heat absorbed by the system is positive, work done by the system is positive, and always subtract work from heat in the First Law equation. Practice identifying whether the system is doing work on the surroundings (expansion) or vice versa (compression).

Question 15

A heat engine absorbs 800 J from a hot reservoir and rejects 300 J to a cold reservoir during one complete cycle. According to the first law of thermodynamics and proper sign conventions, what work is done and what is the change in internal energy for this cyclic process?

  1. W = +500 J done by engine, ΔU = 0 J because it's a complete cycle (correct answer)
  2. W = -500 J done on engine, ΔU = 0 J because it's a complete cycle
  3. W = +500 J done by engine, ΔU = +500 J stored in the engine
  4. W = +800 J done by engine, ΔU = -300 J lost from the engine
  5. W = +200 J done by engine, ΔU = +300 J gained by the engine
Explanation: When analyzing heat engine problems, you need to apply the first law of thermodynamics while carefully tracking energy flows and understanding what happens during complete cycles. The first law states: ΔU=Q−W\Delta U = Q - W, where Q is heat absorbed by the system and W is work done by the system. For this engine, it absorbs 800 J from the hot reservoir and rejects 300 J to the cold reservoir, giving a net heat input of Q=800−300=500 JQ = 800 - 300 = 500 \text{ J}. Since this is a complete cycle, the engine returns to its initial state, meaning the internal energy change must be zero: ΔU=0\Delta U = 0. Using the first law: 0=500−W0 = 500 - W, so W=+500 JW = +500 \text{ J}. The positive sign indicates work done by the engine (expanding and doing work on the surroundings). Choice A correctly identifies both values. Choice B incorrectly shows negative work, which would mean work is done on the engine rather than by it—contradicting how heat engines operate. Choice C correctly calculates the work but wrongly assumes internal energy increases by 500 J, violating the cyclic process requirement that ΔU = 0. Choice D confuses the total heat absorbed (800 J) with the net work output and incorrectly assigns a negative internal energy change. Remember this key principle: for any complete thermodynamic cycle, the internal energy change is always zero because the system returns to its starting point. This makes heat engine problems more straightforward—focus on the net heat flow to find the work output.

Question 16

A refrigerator removes 600 J of heat from its interior and rejects 850 J of heat to the surroundings. From the perspective of the refrigerator as the thermodynamic system, what is the work input and the change in internal energy for one complete cycle?

  1. W = -250 J (work done on refrigerator), ΔU = 0 J (complete cycle) (correct answer)
  2. W = +250 J (work done by refrigerator), ΔU = 0 J (complete cycle)
  3. W = -850 J (work done on refrigerator), ΔU = +600 J (energy gained)
  4. W = +600 J (work done by refrigerator), ΔU = -250 J (energy lost)
  5. W = -250 J (work done on refrigerator), ΔU = +250 J (energy stored)
Explanation: When analyzing refrigerator thermodynamics, you need to apply the first law of thermodynamics while carefully tracking energy flows and sign conventions from the system's perspective. For a refrigerator system, heat flows in two directions: Qc=600 JQ_c = 600 \text{ J} is removed from the cold reservoir (positive because heat enters the system), and Qh=850 JQ_h = 850 \text{ J} is rejected to the hot reservoir (negative because heat leaves the system). Using the first law: ΔU=Q−W\Delta U = Q - W, where Q=Qc−Qh=600−850=−250 JQ = Q_c - Q_h = 600 - 850 = -250 \text{ J}. For a complete cycle, ΔU=0\Delta U = 0, so 0=−250−W0 = -250 - W, giving W=−250 JW = -250 \text{ J}. The negative sign indicates work is done on the refrigerator (as expected, since refrigerators require energy input). Option A correctly identifies both the work input of -250 J and zero internal energy change for a complete cycle. Option B has the wrong sign for work—refrigerators don't do work on their surroundings, they require work input. Option C incorrectly uses 850 J as the work (confusing heat rejection with work) and shows a net internal energy gain, which violates the cyclic process requirement. Option D mistakes the heat removed (600 J) as work output and shows a net energy loss, again violating the cyclic nature. Remember: for any complete thermodynamic cycle, internal energy change is always zero, and refrigerators always require work input (negative W) to move heat from cold to hot reservoirs.

Question 17

Two identical gas samples undergo different processes. Sample A is compressed adiabatically with 200 J200 \text{ J} of work done on it. Sample B undergoes an isothermal compression where 200 J200 \text{ J} of work is also done on it. Which statement correctly describes the heat transfer and internal energy changes?

  1. Both samples have ΔU=+200 J\Delta U = +200 \text{ J}; sample A has Q=0Q = 0, sample B has Q=−200 JQ = -200 \text{ J}
  2. Sample A has ΔU=+200 J\Delta U = +200 \text{ J} and Q=0Q = 0; sample B has ΔU=0\Delta U = 0 and Q=−200 JQ = -200 \text{ J} (correct answer)
  3. Both samples have Q=−200 JQ = -200 \text{ J}; sample A has ΔU=0\Delta U = 0, sample B has ΔU=+200 J\Delta U = +200 \text{ J}
  4. Sample A has ΔU=0\Delta U = 0 and Q=−200 JQ = -200 \text{ J}; sample B has ΔU=+200 J\Delta U = +200 \text{ J} and Q=0Q = 0
Explanation: For both processes, W=−200 JW = -200 \text{ J} (work done on gas is negative work done by gas). Sample A (adiabatic): Q=0Q = 0, so ΔU=Q−W=0−(−200)=+200 J\Delta U = Q - W = 0 - (-200) = +200 \text{ J}. Sample B (isothermal): ΔU=0\Delta U = 0 for ideal gas, so Q=ΔU+W=0+(−200)=−200 JQ = \Delta U + W = 0 + (-200) = -200 \text{ J}. Choice A incorrectly gives both samples the same ΔU\Delta U. Choice C reverses the processes. Choice D completely reverses both QQ and ΔU\Delta U for both processes.

Question 18

A thermodynamic system has work done on it by three different mechanisms simultaneously: a piston compresses the gas (150 J150 \text{ J}), electrical heating performs 75 J75 \text{ J} of work, and magnetic forces do 25 J25 \text{ J} of work on the system. The system also transfers 180 J180 \text{ J} of heat to its surroundings. What is the change in internal energy, and what is the most common sign error students make with this type of problem?

  1. ΔU=+70 J\Delta U = +70 \text{ J}; students often count electrical work as heat instead of work
  2. ΔU=−430 J\Delta U = -430 \text{ J}; students often add all energy transfers without considering directions
  3. ΔU=+70 J\Delta U = +70 \text{ J}; students often use wrong signs for work done on vs by system (correct answer)
  4. ΔU=+430 J\Delta U = +430 \text{ J}; students often forget to account for multiple work contributions separately
Explanation: Total work done on system = 150+75+25=250 J150 + 75 + 25 = 250 \text{ J}, so Wbysystem=−250 JW_{by system} = -250 \text{ J}. Heat transferred to surroundings means Q=−180 JQ = -180 \text{ J}. First law: ΔU=Q−Wby=(−180)−(−250)=+70 J\Delta U = Q - W_{by} = (-180) - (-250) = +70 \text{ J}. The most common error is sign confusion between work done on vs by the system. Choice A incorrectly suggests electrical work isn't work. Choice C gives wrong value. Choice D gives wrong value and misidentifies the common error.

Question 19

A researcher measures the following data for a gas sample undergoing a complex thermodynamic process: Initial state: Ui=500 JU_i = 500 \text{ J}, Final state: Uf=650 JU_f = 650 \text{ J}, Heat interactions: The gas absorbs 200 J200 \text{ J} from reservoir X and simultaneously releases 50 J50 \text{ J} to reservoir Y.

Based on the data provided, determine the work interaction and identify which statement correctly describes both the work value and the most subtle sign convention issue in this analysis.

  1. Work = 0 J0 \text{ J}; the subtle issue is distinguishing between simultaneous heat transfers to different reservoirs (correct answer)
  2. Work = +150 J+150 \text{ J} done by gas; the subtle issue is that internal energy change appears independent of process path
  3. Work = −150 J-150 \text{ J} done by gas; the subtle issue is correctly handling signs when multiple reservoirs interact simultaneously
  4. Work = 0 J0 \text{ J}; the subtle issue is that heat absorbed and released don't obviously relate to internal energy change
Explanation: ΔU=Uf−Ui=650−500=+150 J\Delta U = U_f - U_i = 650 - 500 = +150 \text{ J}. Net heat: Qnet=(+200)+(−50)=+150 JQ_{net} = (+200) + (-50) = +150 \text{ J}. First law: W=Q−ΔU=150−150=0 JW = Q - \Delta U = 150 - 150 = 0 \text{ J}. The subtle issue is that students often struggle with sign conventions when multiple reservoirs are involved simultaneously, wondering whether to treat each heat transfer separately or combine them first. Choice B has wrong work value and misses the sign convention subtlety. Choice C has wrong work value. Choice D correctly identifies zero work but misses the key subtlety about multiple simultaneous heat reservoirs.

Question 20

A gas undergoes a cyclic process where it first expands at constant pressure while absorbing 300 J300 \text{ J} of heat, then is compressed adiabatically back to its initial state. During the isobaric expansion, the gas does 120 J120 \text{ J} of work on its surroundings. What is the work done by the gas during the adiabatic compression?

  1. −120 J-120 \text{ J}
  2. −180 J-180 \text{ J} (correct answer)
  3. +120 J+120 \text{ J}
  4. +180 J+180 \text{ J}
Explanation: For the complete cycle, ΔUcycle=0\Delta U_{cycle} = 0. During isobaric expansion: Q1=300 JQ_1 = 300 \text{ J}, W1=+120 JW_1 = +120 \text{ J}, so ΔU1=Q1−W1=180 J\Delta U_1 = Q_1 - W_1 = 180 \text{ J}. During adiabatic compression: Q2=0Q_2 = 0, and since ΔUcycle=ΔU1+ΔU2=0\Delta U_{cycle} = \Delta U_1 + \Delta U_2 = 0, we have ΔU2=−180 J\Delta U_2 = -180 \text{ J}. Therefore W2=Q2−ΔU2=0−(−180)=−180 JW_2 = Q_2 - \Delta U_2 = 0 - (-180) = -180 \text{ J}. Choice A incorrectly assumes work magnitudes are equal. Choices C and D have wrong signs for compression work.