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Heat and work as the two ways energy crosses a boundary, the sign convention, work at constant pressure, and adiabatic and isothermal processes.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to sign heat and work correctly and use $\Delta U = Q + W$ to track a gas's internal energy and temperature.
From Physics 1 you know that work is force times displacement and that energy is conserved for a system you name. From the last lesson you know that a gas's internal energy is the total energy of its molecules and depends on temperature. This lesson admits heat as a second way energy crosses a boundary.
| Term | What it means |
|---|---|
| First law | $\Delta U = Q + W$: internal energy changes by heat added plus work done on the system. |
| Heat, Q | Energy transferred by a temperature difference; positive when added to the system. |
| Work on the gas, W | Energy transferred when the boundary moves; $W = -P\Delta V$ at constant pressure. |
| Isothermal | At constant temperature; for an ideal gas $\Delta U = 0$. |
| Adiabatic | With no heat transfer, $Q = 0$. |
| Isochoric | At constant volume, so no work is done. |
| Isobaric | At constant pressure. |
Draw a boundary around a gas. Its internal energy $U$ can change only if energy crosses the boundary, and it can cross in two ways: as heat $Q$, flowing because of a temperature difference, and as work $W$, when a force moves the boundary. The first law of thermodynamics is energy conservation with both counted:
$$\Delta U = Q + W,$$
where $Q$ is heat added to the gas and $W$ is work done on the gas, the sign convention of the AP Physics 2 exam. A gas that expands at constant pressure pushes its surroundings and does work $P\Delta V$ on them, so the work done on it is $W = -P\Delta V$.
Another way: picture
Picture a bank account. Deposits come in two forms, paychecks and gifts, and withdrawals too. The balance changes by the total of both. Internal energy is the balance; heat and work are the deposits and withdrawals. It makes no sense to ask how much of the balance is paycheck and how much is gift, just as a gas contains no heat or work, only internal energy.
Another way: steps
Heat flows whenever a system and its surroundings are at different temperatures. Put a pot on a stove and energy flows in; set it on a cold counter and energy flows out. No force needs to move anything.
Work, by contrast, requires a force acting through a displacement. When a gas pushes a piston outward, it exerts a force $PA$ over a distance $d$, doing work $PAd = P\Delta V$ on the piston. When the piston is pushed in, the surroundings do work on the gas. Both change the gas's internal energy, and the first law adds them up.
Physicists and chemists write the first law in two ways. The AP Physics 2 exam, like chemistry, uses $\Delta U = Q + W$ with $W$ the work done on the gas. Engineers often write $\Delta U = Q - W$ with $W$ the work done by the gas. Both say the same thing: energy in raises $U$, energy out lowers it.
The safe habit is to ask of every transfer whether it carries energy into the gas or out. Heat absorbed: in. Heat released: out. Gas compressed: work in. Gas expanding: work out. Then add the ins and subtract the outs.
When pressure stays constant, the work done by a gas is simply $P\Delta V$. With pressure in kilopascals and volume in liters, the product is in joules directly, since $1$ kPa × $1$ L $= 1000$ Pa × $0.001$ m³ $= 1$ J.
A gas at $200$ kPa expanding by $3$ L does $600$ J of work. If it absorbed $1500$ J of heat meanwhile, the first law says its internal energy rose by $900$ J: most of the heat stayed, and the rest left as work. For a monatomic ideal gas at constant pressure, exactly three-fifths of the heat stays and two-fifths becomes work.
If a gas is compressed so quickly that heat has no time to leave, or it is insulated, then $Q = 0$ and all the work done on it becomes internal energy. Its temperature rises. Pump up a bicycle tire vigorously and the pump's barrel grows warm.
The reverse also happens. When a gas expands without heat flowing in, it does work at the expense of its internal energy and cools. Air rising in the atmosphere expands as the pressure falls and cools by about $10$ °C per kilometer, which is why mountaintops are cold and why rising air forms clouds.
Checking an answer. An isothermal process must give $\Delta U = 0$. An adiabatic compression must warm the gas. A gas that absorbs heat and does no work must warm up.
The first law is energy conservation, tested in every experiment ever done. What makes it useful is that internal energy is a property of the state: it depends only on the gas's present condition, not how it got there. Heat and work, by contrast, depend on the path taken.
For an ideal gas, internal energy depends only on temperature, because the molecules exert no forces on each other and all their energy is kinetic. That is why an isothermal process has $\Delta U = 0$ and why a temperature change fixes $\Delta U$ however the gas got there.
Until the 1840s, heat was thought to be a fluid, caloric, that flowed from hot to cold. James Joule, a brewer in Manchester, England, showed that stirring water with a paddle driven by falling weights warmed it by an amount proportional to the work done. Work and heat were two forms of the same thing.
His measurement, $4.18$ joules per calorie, established the first law. The unit of energy bears his name. American nutrition labels still use the food Calorie, a thousand calories, equal to $4184$ J, a direct legacy of Joule's experiments.
A human body is a thermodynamic system. Food supplies internal energy; the body loses it as heat through skin and breath and as work done on the surroundings. At rest, an adult gives off about $100$ W of heat, as much as an old incandescent light bulb.
During exercise, muscles convert internal energy to work with an efficiency of only about $25$ percent. A cyclist producing $150$ W of mechanical power releases about $450$ W of heat at the same time, which is why exercise makes you hot and why gyms run strong air conditioning.
A refrigerator moves heat from its cold interior to the warmer kitchen, the opposite of the natural direction. It does so by compressing a refrigerant gas, which warms it adiabatically above room temperature so it can release heat through the coils at the back. The refrigerant then expands, cooling below the interior temperature so it can absorb heat inside.
Each step is an application of the first law. The compressor's work, from the electric outlet, is the price of moving heat uphill. Air conditioners and heat pumps, now common in American homes, work the same way. A heat pump in winter simply runs the cycle in the other direction, drawing heat from cold outdoor air and releasing it indoors, and can deliver three joules of heat for every joule of electricity it uses.
Imagine taking a gas from one state to another by two different routes: heating at constant volume, then expanding at constant pressure, or the other way round. The work and heat differ between the routes, because the gas expands at different pressures. But the change in internal energy is the same, because both routes start and end at the same temperatures.
This is why internal energy is called a state function and heat and work are not. It is also why a gas taken around a closed cycle, back to its starting state, has $\Delta U = 0$ for the cycle: whatever heat it absorbed overall, it must have given out as the same amount of work. That is the principle of every heat engine, the next lessons' topic.
Diesel engines, which power most American trucks, trains and farm equipment, have no spark plugs. Instead, the piston compresses air to about a twentieth of its volume so quickly that little heat escapes. The work done on the air raises its internal energy, and its temperature climbs above $500$ °C.
Diesel fuel sprayed into this hot air ignites on its own. Because diesel engines compress more than gasoline engines, they convert a larger fraction of fuel energy to work, about $40$ percent against $30$, which is why long-haul trucks use them. The same adiabatic heating explains why a hand-pumped tire valve feels warm after inflating.
A cyclist in a spin class pedaling at $150$ W for thirty minutes does $270$ kJ of work on the bike. Because muscles are only about $25$ percent efficient, her body also releases about $810$ kJ of heat. By the first law, her internal energy falls by $1080$ kJ, about $258$ food Calories.
That heat must leave the body, mostly by sweating, or her temperature would climb dangerously. Evaporating a liter of sweat removes about $2400$ kJ, so a hard hour of riding can cost more than a liter of water. Gyms run fans and air conditioning to help carry the heat away, and coaches urge athletes to drink before they feel thirsty.
A common mistake is to speak of the heat in an object, as if heat were stored. An object stores internal energy; heat and work are only names for energy on the move across a boundary. Once transferred, the energy is simply internal energy, with no memory of how it arrived.
A related error is to add the work done by a gas instead of subtracting it. When a gas expands and pushes on its surroundings, energy leaves it, so its internal energy falls unless heat replaces what the work carried out.
A gas at $150$ kPa absorbs $2400$ J while expanding from $4.0$ L to $10.0$ L. Find the volume change.
$\Delta V = 10.0 - 4.0 = 6.0\ \text{L}$
Final minus initial.
Find the work done by the gas.
$W_{\text{by}} = 150 \times 6.0 = 900\ \text{J}$
kPa times L is J.
Sign the work done on the gas.
$W = -900\ \text{J}$
The gas pushed outward.
Apply the first law.
$\Delta U = 2400 + (-900) = 1500\ \text{J}$
Heat in plus work on.
Interpret the result.
$\Delta U > 0 \Rightarrow \text{the gas warmed}$
Most of the heat stayed.
A bicycle pump compresses $0.020$ mol of air quickly, doing $60$ J of work on it. Note the heat.
$Q \approx 0$
Too quick for heat to escape.
Apply the first law.
$\Delta U = 0 + 60 = 60\ \text{J}$
All work becomes internal energy.
Relate to temperature for air.
$\Delta U = 2.5nR\,\Delta T$
Air molecules also rotate.
Solve for the temperature rise.
$\Delta T = \dfrac{60}{2.5 \times 0.020 \times 8.314} = 144\ \text{K}$
A large warming.
Explain why the pump barrel gets warm.
$\text{hot air heats the metal}$
Heat then leaks out slowly.
Note the connection to engines.
$\text{diesel engines ignite fuel this way}$
Compression alone heats air past $500$ °C.
A gas expands slowly at constant temperature, doing $800$ J of work on a piston. State the internal energy change.
$\Delta U = 0$
Ideal gas at fixed $T$.
Sign the work done on the gas.
$W = -800\ \text{J}$
The gas did the work.
Solve the first law for the heat.
$Q = \Delta U - W = 0 - (-800)$
Rearranged.
Evaluate the heat.
$Q = 800\ \text{J}$
Heat flowed in.
Interpret the balance.
$\text{heat in} = \text{work out}$
The gas passes energy straight through.
Explain why it must be slow.
$\text{time for heat to flow in}$
Otherwise the gas would cool.
Sign the heat.
$Q = -300\ \text{J}$
Released, so negative.
Sign the work.
$W = +500\ \text{J}$
Done on the gas.
Apply the first law.
A gas absorbs $250$ J of heat while it pushes a piston outward, doing $400$ J of work on its surroundings. What is the change in its internal energy, in J?
Complete the worked solution: $1.0$ mol of helium at $300$ K absorbs $3000$ J of heat and does $1200$ J of work pushing a piston. With $R = 8.314$ J/(mol·K), find the change in internal energy in J, the temperature change in K, and the final temperature in K.
Apply the first law.
$\Delta U = Q - W_{\text{by}} =$ u
Heat in minus work out.
Convert to a temperature change.
$\Delta T = \dfrac{\Delta U}{1.5 \times 8.314} =$ d
One mole of a monatomic gas.
Add to the starting temperature.
$T_2 = 300 + \Delta T =$ t
Final temperature.
Check the direction.
$Q > W_{\text{by}} \Rightarrow \text{warmer}$
More energy came in than left.
Match each process to what the first law says about it.
| the internal energy does not change | no heat crosses the boundary | no work is done | the work done by the gas is PΔV | |
|---|---|---|---|---|
| isothermal | ||||
| adiabatic | ||||
| constant volume | ||||
| constant pressure |
A gas at a constant $200$ kPa absorbs $1500$ J of heat and expands by $3$ L. Fill in the work done by the gas, the work done on the gas, and the change in its internal energy, all in J.
| value | |
|---|---|
| work done by the gas (J) | |
| work done on the gas (J) | |
| change in internal energy (J) |
A gas held at a constant $150$ kPa expands by $6$ L while absorbing heat $Q$. Write its change in internal energy, in J, as a function of $Q$ in J.
Answer:
A piston rapidly compresses $0.25$ mol of helium, doing $400$ J of work on it, so quickly that no heat escapes. By how many kelvins does the helium's temperature rise? Use $R = 8.314$ J/(mol·K).
Answer: K
A cyclist on a gym's stationary bike does work on the pedals at $150$ W while her body gives off heat at $450$ W. She rides for $20$ minutes. How many food Calories of her internal energy are used? One food Calorie is $4184$ J.
Answer: Calories
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A gas held at a constant $150$ kPa expands by $6$ L while absorbing heat $Q$. Write its change in internal energy, in J, as a function of $Q$ in J.
Answer:
You can apply the first law. Explain to someone why a gas that absorbs heat can still cool down.
21. Your turn: a gas releases $300$ J of heat while being compressed by $500$ J of work. What is $\Delta U$?, step 3
$\Delta U = -300 + 500 = 200\ \text{J}$
The gas warms slightly.