The First Law of Thermodynamics - AP Physics 2

A fixed amount of gas in a piston-cylinder is the system. Use $\Delta U=Q-W$ , with $Q>0$ into the system and $W>0$ done by the system. During a process, $Q=+90\ \text{J}$ and $\Delta U=-30\ \text{J}$ . Which statement correctly describes the work $W$ done by the gas?
Answer: The gas does $+120\ \text{J}$ of work. This problem involves the first law of thermodynamics. The first law states ΔU = Q - W, which can be rearranged to find work: W = Q - ΔU. Given Q = +90 J (heat added) and ΔU = -30 J (internal energy decreases), we calculate W = 90 - (-30) = 90 + 30 = +120 J. The gas does 120 J of work. Choice A incorrectly subtracts instead of accounting for the negative ΔU, missing that a decrease in internal energy contributes to the work output. When solving for W, always rearrange the first law equation carefully and watch the signs.
A sealed piston-cylinder contains a gas (the system). Use $\Delta U=Q-W$ with $Q>0$ into the gas and $W>0$ done by the gas. In one process, $\Delta U=-120\ \text{J}$ while $Q=+50\ \text{J}$ . Which statement correctly describes the work $W$ ?
Answer: $W=+170\ \text{J}$ because more energy left as work than entered as heat This problem requires applying the first law of thermodynamics. The first law ΔU = Q - W can be rearranged to find work: W = Q - ΔU. Given Q = +50 J (heat added) and ΔU = -120 J (internal energy decreases), we calculate W = 50 J - (-120 J) = 50 J + 120 J = +170 J. The gas does 170 J of work, which explains why internal energy decreases despite heat input—more energy leaves as work than enters as heat. Choice C incorrectly subtracts the internal energy change instead of adding it, showing confusion about sign conventions in the rearranged equation. Always check that your answer makes physical sense: here, work output exceeds heat input, causing the internal energy decrease.
A gas in a piston-cylinder is the system. Use $\Delta U=Q-W$ with $Q>0$ into the gas and $W>0$ done by the gas. During a compression, $W=-300\ \text{J}$ and $\Delta U=+100\ \text{J}$ . Which statement correctly describes the heat transfer $Q$ ?
Answer: $Q=-200\ \text{J}$ because $Q=\Delta U+W=100+(-300)$ This problem requires applying the first law of thermodynamics. The first law ΔU = Q - W can be rearranged to Q = ΔU + W to find heat transfer. Given ΔU = +100 J and W = -300 J (negative because work is done ON the gas during compression), we calculate Q = 100 J + (-300 J) = -200 J. Heat leaves the system because the work done on the gas exceeds the internal energy increase. Choice C makes a sign error by treating the negative work as positive in the calculation, demonstrating confusion about work sign conventions. Always verify your signs: negative Q means heat leaves the system.
A gas (system = gas) is heated in a cylinder. Sign convention: $Q>0$ into system, $W>0$ done by system. During the process, $Q=+90\,\text{J}$ and the internal energy change is $\Delta U=+150\,\text{J}$ . Which statement correctly describes the work $W$ ?
Answer: $W=-60\,\text{J}$ because $W=Q-\Delta U$ This problem involves the first law of thermodynamics. Given Q = +90 J and ΔU = +150 J, we solve for W using ΔU = Q - W. Rearranging: W = Q - ΔU = 90 J - 150 J = -60 J. The negative work means work was done ON the gas (compression). Choice A incorrectly assumes heating always causes expansion. Choice C makes an arithmetic error. Choice D incorrectly claims no work is possible with positive heat transfer. Always verify your result: if internal energy increases more than heat added, work must be done on the system.
A sealed piston-cylinder contains a gas (system = gas). Sign convention: $Q>0$ into system, $W>0$ done by system. In one step, $600\,\text{J}$ of heat enters the gas, and the internal energy increases by $450\,\text{J}$ . Which statement correctly describes the work done by the gas?
Answer: $W=+150\,\text{J}$ because $W=Q-\Delta U$ This problem tests the first law of thermodynamics. Given Q = +600 J (heat enters) and ΔU = +450 J, we find W using ΔU = Q - W. Rearranging: W = Q - ΔU = 600 J - 450 J = +150 J. The positive work means the gas expands and does work on its surroundings. Choice B incorrectly assumes increasing internal energy requires work on the gas. Choice C adds instead of subtracting. Choice D ignores the heat term. Always verify energy conservation: 600 J enters as heat, 450 J increases internal energy, and 150 J leaves as work.
A gas in a cylinder with a movable piston is the system. Use $\Delta U=Q-W$ ( $Q>0$ into gas, $W>0$ by gas). During a compression, an external agent does $300\ \text{J}$ of work on the gas, so $W=-300\ \text{J}$ . At the same time, $Q=-50\ \text{J}$ (heat leaves the gas). Which statement correctly describes $\Delta U$ ?
Answer: The gas's internal energy increases by $250\ \text{J}$ . This problem involves the first law of thermodynamics. The first law states ΔU = Q - W, where positive Q means heat enters the system and positive W means work is done by the system. When work is done ON the gas during compression, W = -300 J (negative because the gas does negative work). With Q = -50 J (heat leaves) and W = -300 J, we calculate ΔU = -50 - (-300) = -50 + 300 = +250 J. The internal energy increases by 250 J. Choice A incorrectly adds the magnitudes instead of using proper signs, a common error when dealing with work done on the system. Always use consistent sign conventions: work BY the system is positive, work ON the system is negative.
Water in a well-insulated container is the system. Use $Q>0$ into the system and $W>0$ done by the system. A paddle wheel driven by an external motor does $300\,\text{J}$ of work on the water (so work done by the system is negative). No heat is exchanged. Which statement correctly describes $\Delta U$ for the water?
Answer: $\Delta U=+300\,\text{J}$ because $Q=0$ and $W=-300\,\text{J}$ so $\Delta U=Q-W$ This problem tests understanding of the first law of thermodynamics. The first law states that the change in internal energy equals heat added to the system minus work done by the system: ΔU = Q - W. The insulated container means Q = 0, and the paddle does 300 J of work ON the water, so W = -300 J (negative because work is done on the system). Substituting: ΔU = 0 - (-300 J) = +300 J. Choice A incorrectly assumes insulation prevents any internal energy change—this is the "insulation blocks all energy" misconception that ignores mechanical work. Always consider both heat AND work as energy transfer mechanisms.
A sealed, rigid metal tank of gas is the system. Take $Q>0$ into the gas and $W>0$ done by the gas. The gas is cooled so that $Q=-500\ \text{J}$ , and because the tank is rigid $W=0$ . Which statement correctly describes $\Delta U$ ?
Answer: $\Delta U=-500\ \text{J}$ because heat leaving reduces internal energy This question tests understanding of the first law of thermodynamics. The first law states ΔU = Q - W. In this rigid tank, W = 0 (no volume change means no work), and Q = -500 J (heat leaves during cooling). Therefore: ΔU = (-500) - 0 = -500 J, meaning internal energy decreases by 500 J. This makes physical sense: cooling removes thermal energy from the gas molecules. Choice B incorrectly claims ΔU = 0, confusing the zero work constraint with zero energy change, a common misconception about rigid containers. For constant volume processes, always remember that ΔU = Q since W = 0.
A gas in a piston is the system. Use $Q>0$ into the gas and $W>0$ done by the gas. In one process, $\Delta U=+100\ \text{J}$ and $Q=-50\ \text{J}$ . Which statement correctly gives the work $W$ done by the gas?
Answer: $W=-150\ \text{J}$ because work must be done on the gas This question tests understanding of the first law of thermodynamics. We need to find W given ΔU = +100 J and Q = -50 J. Using the first law: ΔU = Q - W, we get 100 = (-50) - W. Solving for W: W = -50 - 100 = -150 J. The negative value means work is done ON the gas (compression), which makes sense because heat is leaving while internal energy increases. Choice A incorrectly calculates W = +150 J, likely using W = ΔU - Q instead of the correct rearrangement, a common algebraic error. When solving for work, carefully rearrange ΔU = Q - W to get W = Q - ΔU before substituting values.
A gas in a piston-cylinder (system = gas) undergoes a process. Sign convention: $Q>0$ into system, $W>0$ done by system. The gas releases $120\,\text{J}$ of heat and $80\,\text{J}$ of work is done on the gas. Which statement correctly describes $\Delta U$ ?
Answer: $\Delta U=-40\,\text{J}$ because $Q=-120\,\text{J}$ and $W=-80\,\text{J}$ This problem requires applying the first law of thermodynamics. The gas releases 120 J of heat, so Q = -120 J (heat leaves the system). Work is done ON the gas (80 J), so W = -80 J in our sign convention. Using ΔU = Q - W, we get ΔU = -120 J - (-80 J) = -120 J + 80 J = -40 J. Choice A incorrectly adds the magnitudes without considering signs. Choice C has the wrong sign entirely. Choice D ignores the work term. Always track energy flow direction: heat out and work done on the system both have negative signs in standard convention.
A fixed amount of gas in a piston-cylinder is the system. Use $\Delta U=Q-W$ ( $Q>0$ into gas, $W>0$ by gas). During a process, the gas releases $75\ \text{J}$ of heat ( $Q=-75\ \text{J}$ ) and does $25\ \text{J}$ of work ( $W=+25\ \text{J}$ ). Which statement correctly describes $\Delta U$ ?
Answer: The gas's internal energy decreases by $100\ \text{J}$ . This problem involves the first law of thermodynamics. The first law states ΔU = Q - W, relating internal energy to heat and work. Given Q = -75 J (heat leaves the gas) and W = +25 J (work done by gas), we calculate ΔU = -75 - 25 = -100 J. The internal energy decreases by 100 J. Choice C incorrectly calculates only -50 J, perhaps by subtracting the magnitudes incorrectly or confusing the sign of work. When both heat leaves the system and work is done by the system, both terms reduce internal energy—always add their effects with proper signs.
A sample of gas in a piston-cylinder is the system. Use $\Delta U=Q-W$ with $Q>0$ into the gas and $W>0$ done by the gas. In a process, the surroundings do $350\ \text{J}$ of work on the gas, and $Q=-100\ \text{J}$ leaves the gas. Which statement correctly describes $\Delta U$ ?
Answer: $\Delta U=+250\ \text{J}$ because compression adds more energy than the heat loss removes This problem requires applying the first law of thermodynamics. The first law relates internal energy change to heat and work: ΔU = Q - W. Here, 350 J of work is done ON the gas, which means W = -350 J (negative because the gas does negative work when compressed). With Q = -100 J (heat leaves the gas), we get ΔU = (-100 J) - (-350 J) = -100 J + 350 J = +250 J. The internal energy increases because the work done on the gas adds more energy than the heat loss removes. Choice A incorrectly treats both terms as reducing internal energy, showing the misconception that compression always decreases temperature. Always use proper sign conventions: work done BY the gas is positive, work done ON the gas is negative.
An ideal-gas sample in a piston-cylinder is the system. Use $\Delta U=Q-W$ with $Q>0$ added to the gas and $W>0$ done by the gas. The gas undergoes a slow process in which $\Delta U=0$ while it does $W=+400\ \text{J}$ . Which statement correctly describes the heat transfer $Q$ ?
Answer: Heat must enter the gas: $Q=+400\ \text{J}$ . This problem tests the first law of thermodynamics. The first law states ΔU = Q - W, relating internal energy change to heat and work. Given ΔU = 0 (no change in internal energy) and W = +400 J (work done by the gas), we can solve for Q: 0 = Q - 400, so Q = +400 J. This means 400 J of heat must enter the gas to maintain constant internal energy while it does work. Choice D incorrectly doubles the heat value, perhaps confusing the relationship between Q and W. For processes with ΔU = 0 (like isothermal processes for ideal gases), always remember that Q = W.
A fixed amount of ideal gas in a sealed cylinder with a movable piston is the system. Use $\Delta U=Q-W$ , where $Q>0$ is heat added to the gas and $W>0$ is work done by the gas. During a slow expansion, $Q=+500\ \text{J}$ and the gas does $W=+200\ \text{J}$ . Which statement correctly describes the internal energy change?
Answer: The gas's internal energy increases by $300\ \text{J}$ . This problem tests understanding of the first law of thermodynamics. The first law states that the change in internal energy equals heat added minus work done by the system: ΔU = Q - W. Given Q = +500 J (heat added to gas) and W = +200 J (work done by gas), we calculate ΔU = 500 - 200 = +300 J. The positive result means the internal energy increases by 300 J. Choice D incorrectly assumes that expansion always means no change in internal energy, which is a common misconception—internal energy depends on the balance of heat and work, not just the direction of volume change. Always apply the first law equation systematically, paying attention to sign conventions.
A gas (system = gas) is cooled while being compressed. Sign convention: $Q>0$ into system, $W>0$ done by system. During the process, $Q=-75\,\text{J}$ and $\Delta U=-25\,\text{J}$ . Which statement correctly describes the work $W$ ?
Answer: $W=-50\,\text{J}$ because $W=Q-\Delta U$ This problem involves the first law of thermodynamics. Given Q = -75 J (cooling) and ΔU = -25 J, we solve for W using ΔU = Q - W. Rearranging: W = Q - ΔU = -75 J - (-25 J) = -75 J + 25 J = -50 J. The negative work confirms compression (work done ON the gas). Choice A has the wrong sign. Choice C incorrectly assumes cooling requires positive work. Choice D claims impossible cancellation. Always check consistency: cooling and compression both tend to decrease volume, so they can occur together.
A sample of gas in a piston is the system. Take $Q>0$ into the gas and $W>0$ done by the gas. The internal energy increases by $\Delta U=+400\ \text{J}$ while the gas does $W=+150\ \text{J}$ . Which statement correctly describes the heat transfer $Q$ ?
Answer: $Q=+550\ \text{J}$ because heat supplies both $\Delta U$ and $W$ This question tests understanding of the first law of thermodynamics. The first law can be rearranged to find heat transfer: Q = ΔU + W. Given that ΔU = +400 J (internal energy increases) and W = +150 J (work done by the gas), we calculate Q = 400 + 150 = +550 J. This positive value means heat must enter the gas to supply both the increase in internal energy and the work output. Choice A incorrectly suggests Q = +250 J, possibly by subtracting W from ΔU instead of adding, misunderstanding that both energy changes require heat input. When solving for any variable in the first law, rearrange the equation algebraically before substituting numbers.
A rigid, sealed tank of gas is the system. Take $Q>0$ into the gas and $W>0$ done by the gas. The tank is heated so that $Q=+1200\,\text{J}$ is transferred to the gas; because the tank is rigid, the boundary does not move. Which statement correctly describes the gas's internal energy change?
Answer: $\Delta U=+1200\,\text{J}$ because $W=0$ and $\Delta U=Q-W$ This problem tests understanding of the first law of thermodynamics. The first law states that the change in internal energy equals heat added to the system minus work done by the system: ΔU = Q - W. For a rigid tank, the boundary cannot move, so no work can be done by or on the gas: W = 0. With Q = +1200 J and W = 0, we get ΔU = (+1200 J) - 0 = +1200 J. Choice A incorrectly claims that constant volume prevents internal energy change—this is the "rigid container" misconception that confuses zero work with zero energy change. Always recognize that in rigid containers, all heat transfer goes directly to internal energy change.
A gas in a piston-cylinder is the system. Use $\Delta U=Q-W$ with $Q>0$ into the gas and $W>0$ done by the gas. During an isothermal expansion, $Q=+600\ \text{J}$ is added. Which statement correctly describes the work done by the gas?
Answer: $W=+600\ \text{J}$ because $\Delta U=0$ for an isothermal ideal-gas process This problem requires applying the first law of thermodynamics. For an ideal gas in an isothermal process, temperature remains constant, which means internal energy doesn't change: ΔU = 0. Using the first law ΔU = Q - W with ΔU = 0 and Q = +600 J, we get 0 = 600 J - W, so W = +600 J. The gas does 600 J of work on its surroundings, using all the heat energy added to maintain constant temperature. Choice A incorrectly assumes no work occurs in isothermal processes, confusing constant temperature with no energy transfer. Always remember that isothermal means ΔU = 0 for ideal gases, not that no energy transfers occur.
A sealed cylinder contains an ideal gas (system = gas). Sign convention: $Q>0$ into system, $W>0$ done by system. During one process, $Q=+250\,\text{J}$ and the gas does $W=+150\,\text{J}$ on the piston. Which statement correctly describes the internal energy change of the gas?
Answer: $\Delta U=+100\,\text{J}$ because $\Delta U=Q-W$ This problem tests understanding of the first law of thermodynamics. The first law states that the change in internal energy equals heat added to the system minus work done by the system: ΔU = Q - W. Given Q = +250 J (heat into the gas) and W = +150 J (work done by the gas), we calculate ΔU = 250 J - 150 J = +100 J. Choice A incorrectly assumes heat added always equals the internal energy change, ignoring work. Choice C makes a sign error and confuses work with temperature change. Choice D ignores the work term entirely. Always apply the complete first law equation ΔU = Q - W with careful attention to sign conventions.
A gas in a piston-cylinder is the system. Use $\Delta U=Q-W$ ( $Q>0$ into gas, $W>0$ by gas). The gas is insulated so $Q=0$ , and it expands doing $W=+150\ \text{J}$ . Which statement correctly describes the internal energy change?
Answer: The gas's internal energy decreases by $150\ \text{J}$ . This problem applies the first law of thermodynamics. The first law states ΔU = Q - W, where Q is heat added and W is work done by the system. For an insulated system (adiabatic process), Q = 0. With Q = 0 and W = +150 J (work done by expanding gas), we get ΔU = 0 - 150 = -150 J. The internal energy decreases by 150 J. Choice C incorrectly assumes that no heat transfer means no change in internal energy, ignoring that work also affects internal energy. In adiabatic processes, always remember that any work done comes entirely from the system's internal energy.

Start this quiz