First Law: Closed Systems - Thermodynamics

1. A closed system undergoes a process where the internal energy increases by 180 kJ. If the system receives 250 kJ of heat, what work interaction occurred?
   Answer: 70 kJ of work done by the system When you encounter energy balance problems in thermodynamics, immediately think of the First Law: $\Delta U = Q - W$, where $\Delta U$ is the change in internal energy, $Q$ is heat added to the system, and $W$ is work done by the system. Given that internal energy increases by 180 kJ and the system receives 250 kJ of heat, you can substitute these values: $180 = 250 - W$. Solving for work: $W = 250 - 180 = 70 \text{ kJ}$. Since $W$ is positive in our sign convention, this represents 70 kJ of work done by the system. Answer A correctly identifies this as 70 kJ of work done by the system. Answer B incorrectly suggests work is done on the system—this would require a negative value of $W$ in our equation, which isn't the case here. Answers C and D both arrive at 430 kJ, which results from incorrectly adding heat and internal energy change ($250 + 180 = 430$) instead of applying the First Law properly. This addition approach ignores the fundamental relationship between these energy quantities. Remember the key sign convention: when using $\Delta U = Q - W$, positive $W$ means work done by the system (system does work on surroundings), while negative $W$ means work done on the system. Always write out the First Law equation first, then substitute your known values—this prevents sign errors and conceptual mistakes that lead to wrong answers like adding energies instead of applying conservation principles.
2. A piston-cylinder assembly contains steam that undergoes a process where 200 kJ of heat is added and the steam does 80 kJ of work. If the mass of steam is 0.5 kg and the specific internal energy increases by 240 kJ/kg, verify whether this process is consistent with the first law of thermodynamics.
   Answer: Yes, the process is consistent with the first law When you encounter a thermodynamics problem involving heat, work, and internal energy changes, you're dealing with the first law of thermodynamics. This fundamental principle states that energy cannot be created or destroyed, only converted from one form to another. For a closed system, the first law is expressed as: $Q = \Delta U + W$, where Q is heat transfer, ΔU is the change in internal energy, and W is work done by the system. Let's verify this process step by step. First, calculate the total change in internal energy: $\Delta U = m \times \Delta u = 0.5 \text{ kg} \times 240 \text{ kJ/kg} = 120 \text{ kJ}$ Now apply the first law equation: $Q = \Delta U + W = 120 \text{ kJ} + 80 \text{ kJ} = 200 \text{ kJ}$ This matches exactly with the given heat addition of 200 kJ, confirming the process is consistent with the first law. Answer A is correct because our calculation confirms the energy balance. Answer B incorrectly suggests heat should be removed rather than added, which would violate the energy balance given the positive internal energy change and work output. Answer C claims 280 kJ should be added, but this would create an energy imbalance of 80 kJ. Answer D suggests different work values, but changing the work would disrupt the established energy balance. Remember: always check thermodynamic processes against the first law by ensuring $Q = \Delta U + W$. This energy balance must hold for any realistic process in a closed system.
3. Two identical closed systems A and B contain the same gas at the same initial state. System A undergoes an isothermal expansion doing 60 kJ of work. System B undergoes an adiabatic expansion doing the same amount of work. Compare the final internal energies of the two systems.
   Answer: System A has 60 kJ higher internal energy than system B When you encounter thermodynamic process comparisons, focus on applying the first law of thermodynamics: $\Delta U = Q - W$, where ΔU is the change in internal energy, Q is heat added, and W is work done by the system. For System A (isothermal process): In an isothermal expansion, temperature remains constant, so ΔU = 0 for an ideal gas (internal energy depends only on temperature). Since the system does 60 kJ of work, the first law gives us: 0 = Q - 60 kJ, so Q = 60 kJ. Heat must flow into the system to maintain constant temperature while doing work. For System B (adiabatic process): No heat exchange occurs (Q = 0). The system does 60 kJ of work, so: ΔU = 0 - 60 kJ = -60 kJ. The internal energy decreases because the system uses its own internal energy to do work, causing temperature to drop. Comparing final internal energies: System A maintains its original internal energy (ΔU = 0), while System B loses 60 kJ of internal energy. Therefore, System A has 60 kJ higher internal energy than System B. Choice A is correct. Choice B reverses the relationship—System B actually has lower internal energy. Choice C ignores that adiabatic work reduces internal energy while isothermal work doesn't. Choice D incorrectly suggests a 120 kJ difference, likely from adding rather than comparing the energy changes. Study tip: Remember that isothermal processes maintain constant internal energy through heat exchange, while adiabatic processes change internal energy directly through work since no heat transfer occurs.
4. A rigid container holds 0.5 kg of an ideal gas. The gas undergoes a process where 150 kJ of heat is added while the gas does 40 kJ of work on its surroundings. If the initial internal energy of the gas is 300 kJ, what is the final internal energy?
   Answer: 410 kJ This question tests your understanding of the First Law of Thermodynamics, which is essentially conservation of energy applied to thermal systems. When you see a problem involving heat transfer, work, and internal energy changes, immediately think of the First Law equation. The First Law states: $\Delta U = Q - W$, where $\Delta U$ is the change in internal energy, $Q$ is heat added to the system, and $W$ is work done by the system. Here, 150 kJ of heat is added ($Q = +150$ kJ) and the gas does 40 kJ of work on its surroundings ($W = +40$ kJ). Calculating the change in internal energy: $\Delta U = 150 - 40 = 110$ kJ. Since the initial internal energy is 300 kJ, the final internal energy is $300 + 110 = 410$ kJ. Choice A (410 kJ) is correct. Choice B (450 kJ) represents the common error of adding both heat and work to the initial internal energy, forgetting that work done by the system reduces internal energy. Choice C (490 kJ) likely comes from incorrectly adding all three values together (300 + 150 + 40). Choice D (190 kJ) results from subtracting both heat and work from the initial internal energy, misunderstanding the signs in the First Law equation. Remember: when applying the First Law, be extremely careful with signs. Heat added to the system increases internal energy, while work done by the system decreases it. Always double-check whether work is done by or on the system.
5. A closed system receives 300 kJ of heat while its volume decreases from 0.1 m³ to 0.06 m³ against a constant external pressure of 500 kPa. What is the change in internal energy of the system?
   Answer: 320 kJ increase When you encounter a thermodynamics problem involving heat transfer and volume changes, you're dealing with the First Law of Thermodynamics: $\Delta U = Q - W$, where $\Delta U$ is the change in internal energy, $Q$ is heat added to the system, and $W$ is work done by the system. First, identify what you know: the system receives 300 kJ of heat ($Q = +300$ kJ), and the volume decreases from 0.1 m³ to 0.06 m³ against constant pressure. For work done by the system against constant external pressure: $W = P \times \Delta V = 500 \text{ kPa} \times (0.06 - 0.1) \text{ m}^3 = 500 \times (-0.04) = -20 \text{ kJ}$ Since work done by the system is negative (compression), the system actually has work done on it. Therefore: $\Delta U = 300 - (-20) = 320 \text{ kJ}$ Answer A (320 kJ increase) is correct. Answer B (280 kJ increase) represents the common error of adding work incorrectly: $300 + (-20) = 280$ kJ, forgetting that work done on the system increases internal energy. Answer C (320 kJ decrease) gets the magnitude right but the wrong sign, possibly from misunderstanding the sign convention. Answer D (280 kJ decrease) combines both sign errors. Remember: when volume decreases against external pressure, work is done on the system (negative work done by the system), which increases internal energy. Always carefully track your signs in the First Law equation.
6. A rigid tank contains air that is heated such that its internal energy increases by 400 kJ while 600 kJ of heat is transferred to the air. A student claims that 200 kJ of work was done by the air during this process. What is wrong with this analysis?
   Answer: No work can be done by the air since the tank is rigid and volume remains constant For a rigid tank, volume is constant, so no boundary work can be performed (W = ∫PdV = 0 when dV = 0). The student's claim of 200 kJ work is impossible. For the actual process: ΔU = Q - W, so 400 = 600 - 0 = 600 kJ, which shows the given values are also inconsistent, but the fundamental error is assuming work can occur in a rigid container. Choice A incorrectly suggests work occurs. Choice C addresses a secondary issue. Choice D miscalculates work.
7. A piston-cylinder device contains 2 kg of steam. The steam receives 800 kJ of heat and its internal energy increases by 600 kJ. If the piston moves against a constant external pressure of 300 kPa, what is the change in volume of the steam?
   Answer: 0.67 m³ increase When you encounter a piston-cylinder problem involving heat transfer and internal energy changes, you're dealing with the first law of thermodynamics. The key relationship here is $Q = \Delta U + W$, where Q is heat added, ΔU is the change in internal energy, and W is work done by the system. Given that the steam receives 800 kJ of heat and its internal energy increases by 600 kJ, you can find the work done by the system: $W = Q - \Delta U = 800 - 600 = 200 \text{ kJ}$. For a piston moving against constant external pressure, the work done by the system is $W = P \Delta V$, where P is the external pressure and ΔV is the volume change. Solving for volume change: $\Delta V = \frac{W}{P} = \frac{200 \text{ kJ}}{300 \text{ kPa}} = \frac{200,000 \text{ J}}{300,000 \text{ Pa}} = 0.67 \text{ m}^3$ Since work is positive (system does work on surroundings), the volume increases by 0.67 m³, making A correct. Option B incorrectly assumes the volume decreases, which would happen if work were done on the system rather than by it. Option C uses the wrong formula, possibly calculating $\Delta V = \frac{Q}{P}$ instead of using the work term. Option D likely adds the heat and internal energy changes before dividing by pressure, representing a fundamental misunderstanding of the first law. Remember: always apply the first law systematically to find work, then use $W = P\Delta V$ for constant pressure processes. The sign of work tells you whether volume increases or decreases.
8. A closed system executes a thermodynamic cycle consisting of three processes. The net work output of the cycle is 250 kJ. During the cycle, the system receives 600 kJ of heat in process A, rejects 200 kJ of heat in process B, and has an unknown heat transfer in process C. What is the heat transfer in process C?
   Answer: 150 kJ is rejected by the system during process C For a complete cycle, ΔU = 0, so Q_net = W_net = 250 kJ. Net heat input = 600 - 200 + $Q_C$ = 250 kJ, so $Q_C$ = -150 kJ (rejected). Choice B has correct magnitude but wrong direction. Choice C incorrectly calculates $Q_C$ = -(600 - 200 + 250). Choice D incorrectly adds all values: 600 + 200 + 250 = 1050, but uses 850.
9. A piston-cylinder assembly contains air that undergoes a process where the pressure remains constant at 250 kPa. The air expands from 0.02 m³ to 0.05 m³ while 60 kJ of heat is added. What is the change in internal energy?
   Answer: 52.5 kJ increase When you encounter a thermodynamics problem involving a piston-cylinder with constant pressure, you're dealing with an isobaric process. The key is applying the first law of thermodynamics: $\Delta U = Q - W$, where $\Delta U$ is the change in internal energy, $Q$ is heat added, and $W$ is work done by the system. For an isobaric process, work is calculated as $W = P \Delta V$. Here, the work done by the expanding gas is: $W = 250 \text{ kPa} \times (0.05 - 0.02) \text{ m}^3 = 250 \times 0.03 = 7.5 \text{ kJ}$ Applying the first law with $Q = 60 \text{ kJ}$ and $W = 7.5 \text{ kJ}$: $\Delta U = 60 - 7.5 = 52.5 \text{ kJ}$ Since this is positive, internal energy increases by 52.5 kJ, making answer A correct. Answer B (67.5 kJ increase) results from incorrectly adding work to heat instead of subtracting it—a sign confusion in the first law equation. Answer C (52.5 kJ decrease) gets the magnitude right but the wrong sign, likely from misunderstanding that positive heat addition typically increases internal energy. Answer D (7.5 kJ increase) confuses the work value with the change in internal energy. Remember: in expansion processes, the system does work on the surroundings, so work is subtracted from the heat added when calculating internal energy change. Always check your sign conventions carefully.
10. A closed system undergoes a thermodynamic process where 800 kJ of heat is added while the system performs 300 kJ of work on the surroundings. If the initial internal energy is 1200 kJ, what is the final internal energy of the system?
    Answer: 1700 kJ Using the first law for closed systems: ΔU = Q - W = 800 kJ - 300 kJ = 500 kJ. Final internal energy = Initial + ΔU = 1200 kJ + 500 kJ = 1700 kJ. Choice B incorrectly subtracts heat instead of work. Choice C incorrectly calculates ΔU as Q + W and subtracts from initial energy. Choice D incorrectly adds both Q and W to initial energy.
11. An ideal gas in a piston-cylinder device expands from 2 m³ to 5 m³ against a constant external pressure of 100 kPa while receiving 800 kJ of heat. If the gas temperature increases such that its internal energy rises by 500 kJ, what work is done by the gas?
    Answer: 300 kJ of work is done by the gas on the surroundings Using the first law: ΔU = Q - W, so 500 = 800 - W, giving W = 300 kJ done by the system. This can be verified: W = P_ext × ΔV = 100 kPa × (5-2) m³ = 300 kJ. Choice B has the correct magnitude but wrong direction. Choice C incorrectly adds Q and ΔU. Choice D incorrectly calculates W = Q - ΔU but assigns wrong direction.
12. An ideal gas undergoes an isobaric process at 400 kPa. The gas volume changes from 0.3 m³ to 0.8 m³, and the internal energy increases by 180 kJ. Determine the heat transfer.
    Answer: 380 kJ added When you encounter an isobaric (constant pressure) process, you need to apply the first law of thermodynamics while accounting for the work done by the expanding gas. The first law states: $Q = \Delta U + W$, where Q is heat transfer, ΔU is the change in internal energy, and W is work done by the system. For an isobaric process, work is calculated as $W = P \Delta V$. Given the pressure P = 400 kPa and volume change from 0.3 m³ to 0.8 m³: $W = 400 \text{ kPa} \times (0.8 - 0.3) \text{ m³} = 400 \times 0.5 = 200 \text{ kJ}$ With ΔU = 180 kJ (given), the heat transfer becomes: $Q = 180 \text{ kJ} + 200 \text{ kJ} = 380 \text{ kJ}$ Since Q is positive, heat is added to the system, making A) 380 kJ added correct. B) 20 kJ removed represents the common error of subtracting work from internal energy (180 - 200 = -20), then incorrectly interpreting the negative as "removed." C) 200 kJ added occurs when students confuse work with total heat transfer, forgetting to add the internal energy change. D) 580 kJ added results from incorrectly adding pressure and volume change to internal energy instead of calculating work properly. Remember: in isobaric expansion problems, always calculate work as P×ΔV first, then add it to the internal energy change. The expanding gas does positive work on its surroundings, requiring additional heat input beyond what's needed for internal energy increase.
13. A gas undergoes an adiabatic expansion in a closed system. The gas does 120 kJ of work on its surroundings. If the initial internal energy is 800 kJ, what is the final internal energy?
    Answer: 680 kJ When you encounter adiabatic processes in thermodynamics, remember that "adiabatic" means no heat transfer occurs (Q = 0). This constraint dramatically simplifies your analysis using the first law of thermodynamics. The first law states: $\Delta U = Q - W$, where ΔU is the change in internal energy, Q is heat added to the system, and W is work done by the system. Since this is an adiabatic process, Q = 0, so the equation becomes: $\Delta U = -W$. Given that the gas does 120 kJ of work on its surroundings, W = +120 kJ (positive because work is done by the system). Therefore: $\Delta U = -120 \text{ kJ}$. With an initial internal energy of 800 kJ, the final internal energy is: $U_{final} = 800 - 120 = 680 \text{ kJ}$. Choice A (680 kJ) is correct. Choice B (920 kJ) incorrectly adds the work to the internal energy, forgetting that when a system does work, it loses energy. Choice C (800 kJ) assumes no change in internal energy, which would only occur if no work were done. Choice D (680 kJ with heat transfer of 120 kJ) reaches the right numerical answer but contradicts the adiabatic condition by suggesting heat transfer occurred. Remember this key relationship: in adiabatic processes, all energy changes come from work alone. When a gas expands adiabatically and does work, its internal energy (and temperature) must decrease since no heat compensates for the energy lost through work.
14. A closed system undergoes a process where its internal energy decreases by 150 kJ. During this process, the system rejects 100 kJ of heat to the surroundings. A student calculates that 50 kJ of work was done by the system. What is the error in this analysis?
    Answer: Work should be 50 kJ done on the system, not by the system, during this process Using the first law: ΔU = Q - W, so -150 = -100 - W, giving W = -100 + 150 = 50 kJ. However, this means 50 kJ of work is done ON the system (negative work in our convention means work done on the system). Choice A incorrectly states direction. Choice B miscalculates as W = Q + ΔU. Choice D incorrectly questions the given values rather than the work direction.
15. During a compression process in a closed system, 25 kJ of work is done on a gas while 15 kJ of heat is removed from the gas. What is the change in internal energy of the gas?
    Answer: 10 kJ increase When you encounter a thermodynamics problem involving work and heat transfer in a closed system, immediately think of the First Law of Thermodynamics: $\Delta U = Q - W$, where $\Delta U$ is the change in internal energy, $Q$ is heat added to the system, and $W$ is work done by the system. The key is establishing the correct sign conventions. Since 15 kJ of heat is removed from the gas, $Q = -15$ kJ (negative because heat leaves the system). Since 25 kJ of work is done on the gas during compression, $W = -25$ kJ (negative because work is done on the system, not by it). Applying the First Law: $\Delta U = Q - W = (-15) - (-25) = -15 + 25 = +10$ kJ. The positive value indicates an increase in internal energy, making A) 10 kJ increase correct. Let's examine why the other options are wrong. Option B) 40 kJ increase results from incorrectly adding the magnitudes: $25 + 15 = 40$, ignoring proper sign conventions. Option C) 10 kJ decrease comes from using $\Delta U = Q + W = -15 + (-25) = -40$ kJ, then taking the wrong magnitude and sign. Option D) 40 kJ decrease represents the same magnitude error as B but with the wrong sign. Study tip: Always establish your sign convention first in thermodynamics problems. Remember that compression work done on a gas and heat removed from a system are both negative in the standard convention, but they affect internal energy in opposite ways according to the First Law.
16. A gas undergoes an adiabatic compression process where 400 kJ of work is done on the gas. If the initial internal energy is 1500 kJ, what can be concluded about the final state of the gas?
    Answer: The final internal energy is 1900 kJ and the gas temperature has increased significantly For an adiabatic process, Q = 0. When work is done ON the gas (compression), this increases the internal energy. Using the first law: ΔU = Q - W. Since 400 kJ is done ON the gas, W = -400 kJ in our sign convention. Therefore: ΔU = 0 - (-400) = 400 kJ. Final internal energy = 1500 + 400 = 1900 kJ. Since internal energy increased, temperature increased significantly. Choice B treats compression as expansion. Choice C ignores work effects. Choice D incorrectly suggests insufficient information.
17. During an isothermal process in a closed system containing an ideal gas, 45 kJ of work is done by the gas. What can be concluded about the heat transfer?
    Answer: 45 kJ of heat is added to the system When you encounter isothermal processes with ideal gases, remember that temperature remains constant throughout. This seemingly simple condition has profound implications for energy analysis. For an isothermal process in an ideal gas, the internal energy change is zero ($\Delta U = 0$) because internal energy depends only on temperature for ideal gases. Applying the first law of thermodynamics: $\Delta U = Q - W$, where Q is heat added to the system and W is work done by the system. Since $\Delta U = 0$, we get $0 = Q - W$, which means $Q = W$. The problem states that 45 kJ of work is done by the gas, so W = +45 kJ. Therefore, Q must also equal +45 kJ, meaning 45 kJ of heat is added to the system. This makes option A correct. Option B represents a sign error—if 45 kJ were removed, the gas couldn't do positive work while maintaining constant temperature. Option C reflects a common misconception that isothermal means "no heat transfer," but isothermal actually means "constant temperature." Heat transfer is essential to maintain that constant temperature as the gas expands and does work. Option D suggests you need pressure data, but the first law of thermodynamics provides all the information needed—pressure data would only help determine specific process details, not the fundamental energy balance. Study tip: For isothermal ideal gas processes, always remember that Q = W. The heat transfer exactly equals the work done to maintain constant temperature, making these problems straightforward once you recognize the constraint.
18. An ideal gas in a rigid container has an initial temperature of 300 K and internal energy of 450 kJ. Heat is added until the temperature reaches 400 K. If the specific heat at constant volume is 0.75 kJ/kg·K and the mass is 2 kg, what is the heat added to the system?
    Answer: 150 kJ When you encounter a rigid container problem in thermodynamics, you're dealing with a constant volume process where no work is done by or on the gas. This means all the heat added goes directly into changing the internal energy of the system. For a constant volume process with an ideal gas, you can calculate heat added using $Q = mc_v\Delta T$, where $m$ is mass, $c_v$ is specific heat at constant volume, and $\Delta T$ is the temperature change. Here, the temperature increases from 300 K to 400 K, so $\Delta T = 100$ K. With $m = 2$ kg and $c_v = 0.75$ kJ/kg·K: $Q = (2 \text{ kg})(0.75 \text{ kJ/kg}\cdot\text{K})(100 \text{ K}) = 150 \text{ kJ}$ This confirms answer A is correct. Answer B (225 kJ) likely comes from incorrectly using the final temperature (400 K) instead of the temperature change: $2 \times 0.75 \times 150 = 225$ kJ. Answer C (75 kJ) results from using only half the mass or half the specific heat. Answer D (300 kJ) might stem from doubling the correct answer or confusing this with the final temperature value. Remember that in rigid container problems, focus on the temperature change, not the absolute temperatures. The key formula is $Q = mc_v\Delta T$ since no work is done when volume is constant. Always double-check that you're using the temperature difference, not individual temperature values.
19. A gas expands in a cylinder against a variable external pressure. The work done by the gas is 75 kJ, and the internal energy decreases by 25 kJ. What is the magnitude and direction of heat transfer?
    Answer: 50 kJ added to the system When you encounter thermodynamics problems involving work, internal energy, and heat, immediately think of the First Law of Thermodynamics: $\Delta U = Q - W$, where $\Delta U$ is the change in internal energy, $Q$ is heat added to the system, and $W$ is work done by the system. Let's identify what we know: the gas does 75 kJ of work ($W = +75$ kJ, positive because work is done by the system), and internal energy decreases by 25 kJ ($\Delta U = -25$ kJ). Substituting into the First Law equation: $-25 = Q - 75$ Solving for Q: $Q = -25 + 75 = +50$ kJ The positive sign means 50 kJ of heat is added to the system, making answer A correct. Let's examine why the other options fail: Answer B (50 kJ removed) would result from incorrectly treating the work as negative or misapplying the sign convention. Answer C (100 kJ added) likely comes from adding the magnitudes (75 + 25) without proper consideration of the First Law. Answer D (100 kJ removed) represents a double error—both incorrect magnitude and wrong direction. The key insight is that even though the gas loses internal energy, it must receive heat to perform the expansion work. Think of it this way: the gas needs energy input to do work, but some of that input goes toward the work itself rather than increasing internal energy. Study tip: Always write out the First Law equation and carefully track signs—positive Q means heat added, positive W means work done by the system.
20. During a constant volume process, 2 kg of air receives 150 kJ of heat. If the specific heat at constant volume is 0.718 kJ/kg·K, what is the change in internal energy?
    Answer: 150 kJ increase When you encounter constant volume thermodynamic processes, remember that the first law of thermodynamics is your foundation: $\Delta U = Q - W$, where $\Delta U$ is the change in internal energy, $Q$ is heat added, and $W$ is work done by the system. The key insight for constant volume processes is that no work is performed because there's no volume change. Since work equals $P\Delta V$, and $\Delta V = 0$, we have $W = 0$. This simplifies the first law to $\Delta U = Q$. Given that 150 kJ of heat is added to the system, the change in internal energy equals exactly 150 kJ. The mass of air (2 kg) and specific heat capacity (0.718 kJ/kg·K) are provided to calculate temperature change if needed, but they're not required for finding $\Delta U$. Answer A (150 kJ increase) is correct because it directly applies the first law with zero work. Answer B (104.5 kJ increase) incorrectly assumes some of the heat energy went into doing work, which is impossible at constant volume. Answer C (208.8 kJ increase) appears to add the heat input to some calculated value using the given specific heat data, showing confusion about which thermodynamic relationship to apply. Answer D (0 kJ) incorrectly assumes that all heat energy somehow converts to work or that internal energy doesn't change when heat is added. Study tip: For constant volume processes, always remember that $W = 0$, so the change in internal energy always equals the heat transferred. This makes these problems more straightforward than other thermodynamic processes.

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