The First Law of Thermodynamics
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AP Physics 2 › The First Law of Thermodynamics
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?
$\Delta U=-1200,\text{J}$ because no work is done so energy must leave as heat
$\Delta U=0,\text{J}$ because the volume is constant so heat cannot change internal energy
$\Delta U=+1200,\text{J}$ because temperature always increases by the same amount as heat added
$\Delta U=+1200,\text{J}$ because $W=0$ and $\Delta U=Q-W$
Explanation
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. Take $Q>0$ into the gas and $W>0$ done by the gas. During a process the gas does $W=+300,\text{J}$ of work, and its internal energy does not change ($\Delta U=0$). Which statement correctly describes the heat transfer?
$Q=0,\text{J}$ because no internal energy change implies no heat transfer
$Q=-300,\text{J}$ because $Q=\Delta U-W$
$Q=+300,\text{J}$ because work done by the gas always increases its temperature
$Q=+300,\text{J}$ because $Q=W$ when $\Delta U=0$
Explanation
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 ΔU = 0 and W = +300 J, we can solve for Q: 0 = Q - (+300 J), which gives Q = +300 J. Choice A incorrectly assumes no internal energy change means no heat transfer—this is the "zero ΔU means isolated system" misconception that ignores the possibility of balanced heat and work. Always recognize that ΔU = 0 means heat added equals work done by the system.
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 $W$?
$W=-170\ \text{J}$ because the gas must have been compressed
$W=+170\ \text{J}$ because more energy left as work than entered as heat
$W=-70\ \text{J}$ because positive heat implies negative work
$W=+70\ \text{J}$ because work equals heat minus internal energy change
Explanation
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.
Liquid water in a rigid, sealed container is the system. Use $\Delta U=Q-W$ with $Q>0$ into the water and $W>0$ done by the water. A heater transfers $Q=+800\ \text{J}$ to the water while the container’s volume stays constant. Which statement correctly describes the energy change of the system?
$\Delta U=0$ because constant volume implies no heat transfer
$\Delta U=+800\ \text{J}$ because $W=0$ in a rigid container
$\Delta U$ cannot be determined because temperature change is not given
$\Delta U=-800\ \text{J}$ because heat added is used entirely for expansion work
Explanation
This problem requires applying the first law of thermodynamics. The first law states ΔU = Q - W, where Q is heat added to the system and W is work done by the system. In a rigid container, the volume cannot change, so no expansion or compression work can occur: W = 0. With Q = +800 J added to the water and W = 0, we get ΔU = 800 J - 0 = +800 J. All the heat energy goes into increasing the internal energy of the water. Choice A incorrectly assumes constant volume prevents heat transfer, confusing the constraint on work with a constraint on heat. Always identify whether the system can do boundary work before applying the first law.
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 $\Delta U$?
$\Delta U=-450\ \text{J}$ because both heat loss and compression reduce $U$
$\Delta U=+100\ \text{J}$ because negative heat means temperature must increase
$\Delta U=+250\ \text{J}$ because compression adds more energy than the heat loss removes
$\Delta U=-100\ \text{J}$ because internal energy change equals heat transfer
Explanation
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.
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?
$\Delta U=-300,\text{J}$ because work done on the system reduces its internal energy
$\Delta U=0,\text{J}$ because insulation prevents any internal energy change
$\Delta U=+300,\text{J}$ because $Q=0$ and $W=-300,\text{J}$ so $\Delta U=Q-W$
$\Delta U=+300,\text{J}$ because any work implies heat is added
Explanation
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 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 $\Delta U$?
The gas’s internal energy decreases by $50\ \text{J}$.
The gas’s internal energy increases by $50\ \text{J}$ because work is done by the gas.
The gas’s internal energy decreases by $100\ \text{J}$.
The gas’s internal energy is unchanged because heat transfer sets temperature only.
Explanation
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 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?
The gas’s internal energy decreases by $150\ \text{J}$ only if temperature decreases.
The gas’s internal energy decreases by $150\ \text{J}$.
The gas’s internal energy is unchanged because no heat is transferred.
The gas’s internal energy increases by $150\ \text{J}$ because it expands.
Explanation
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.
A gas in a sealed cylinder with a movable piston is the system. Use $\Delta U=Q-W$ ($Q>0$ into gas, $W>0$ by gas). During a process, the gas does $W=+80\ \text{J}$, and its internal energy increases by $\Delta U=+20\ \text{J}$. Which statement correctly describes the heat transfer $Q$?
Heat leaves the gas: $Q=-60\ \text{J}$.
No heat is transferred: $Q=0\ \text{J}$.
Heat enters the gas: $Q=+100\ \text{J}$.
Heat enters the gas: $Q=+60\ \text{J}$.
Explanation
This problem tests the first law of thermodynamics. The first law states ΔU = Q - W, which rearranges to Q = ΔU + W for finding heat transfer. Given W = +80 J (work done by gas) and ΔU = +20 J (internal energy increases), we calculate Q = 20 + 80 = +100 J. Heat enters the gas. Choice C gives only 60 J, perhaps by subtracting W from ΔU instead of adding, a sign error when rearranging the equation. Always solve for the unknown by properly rearranging the first law equation and maintaining sign consistency.
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?
The gas does $+60\ \text{J}$ of work.
An external agent does $+60\ \text{J}$ of work on the gas.
The gas does $+120\ \text{J}$ of work.
Work is zero because internal energy decreases.
Explanation
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.