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Compression work

The work done on a gas as $-\int P\,dV$ along constant-pressure, isothermal and adiabatic paths, the adiabatic relations $PV^\gamma$ and $TV^{\gamma - 1}$, and the heat from the first law.

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.

1. What you will learn

By the end of this lesson you will be able to find the work done on a gas along an isobaric, isothermal or adiabatic path with its sign, derive and use the adiabatic relations, find the heat in each process from the first law, and explain why work and heat depend on the path.

2. What you already have

You know the first law as energy accounting, that an ideal gas holds thermal energy $U = \tfrac{f}{2}nRT$, and from Physics 2 that the work in a process is the area under its path on a pressure-volume diagram. This lesson turns that area into an integral and evaluates it for the paths a gas actually follows, including one along which no heat flows at all.

3. Words for this lesson

4. Work is the area under the path

Push a piston of area $A$ a distance $dx$ into a cylinder of gas at pressure $P$. The force is $PA$, the work done on the gas is $PA\,dx$, and $A\,dx$ is the decrease in volume, $-dV$. So the work done on a gas in a small step is $-P\,dV$, and for a whole process

$$W = -\int_{V_i}^{V_f} P\,dV.$$

Compression ($dV < 0$) does positive work on the gas; expansion does negative work, because then the gas is doing work on whatever it pushes. The first law, $\Delta U = Q + W$, then tells how much heat must have crossed the boundary.

The integral needs $P$ as a function of $V$, so it depends on the path, and three paths do most of the work in thermal physics.

• Constant pressure: $W = -P(V_f - V_i)$.
• Constant temperature: $P = nRT/V$, so $W = -nRT\int dV/V = nRT\ln(V_i/V_f)$. Since $U$ does not change, all of that work leaves as heat: $Q = -W$.
• No heat (adiabatic): all the work stays in the gas, $dU = -P\,dV$. With $U = \tfrac{f}{2}NkT$ and $P = NkT/V$ this becomes $\tfrac{f}{2}\,dT/T = -dV/V$, which integrates to

$$TV^{\gamma - 1} = \text{constant}, \qquad PV^{\gamma} = \text{constant}, \qquad \gamma = \frac{f + 2}{f}.$$

An adiabat on a $PV$ diagram is steeper than an isotherm through the same point, because as the gas is squeezed it also heats, and both effects raise the pressure.

5. Deriving the adiabat

The adiabatic relations are worth deriving once, because the derivation shows where $\gamma$ comes from. With $Q = 0$ the first law is $dU = W$, and for a quasistatic step $W = -P\,dV$:

$$\tfrac{f}{2}Nk\,dT = -P\,dV = -\frac{NkT}{V}\,dV.$$

Divide both sides by $NkT$ and separate:

$$\frac{f}{2}\frac{dT}{T} = -\frac{dV}{V}.$$

Integrate from the start to the end of the process:

$$\frac{f}{2}\ln\frac{T_f}{T_i} = -\ln\frac{V_f}{V_i} \quad\Rightarrow\quad T_fV_f^{2/f} = T_iV_i^{2/f}.$$

Since $\gamma = (f + 2)/f$, the exponent $2/f$ is $\gamma - 1$, so $TV^{\gamma - 1}$ is constant. Replacing $T$ by $PV/Nk$ gives $PV^{\gamma}$ constant. For air, $f = 5$ and $\gamma = 1.4$; for helium, $f = 3$ and $\gamma = 5/3$. A gas with more degrees of freedom has a smaller $\gamma$: its energy is spread over more modes, so a given amount of work raises its temperature less.

Two conditions are hidden in this derivation. The process must be quasistatic, so that $P = NkT/V$ holds at every moment, and the gas must be ideal. A gas that rushes into a vacuum through a burst membrane has no heat and no work, and its temperature does not change at all: it is adiabatic but not quasistatic, and the adiabat does not describe it.

6. The method, step by step, and how to check it

1. Draw the path on a $PV$ diagram, even roughly. Compression runs to the left, expansion to the right; the area under the path is the size of the work.
2. Identify the process by what the problem holds fixed. A water bath or slow change: isothermal. Insulated walls or a fast stroke: adiabatic. A weighted piston free to move: isobaric. A rigid tank: no work.
3. Write the work with its sign: $-P\Delta V$, $nRT\ln(V_i/V_f)$, or, for an adiabat, $W = \Delta U = \tfrac{f}{2}nR(T_f - T_i)$, which is easier than integrating $PV^{\gamma}$.
4. Use consistent units. Kilopascals times litres give joules directly; pascals times cubic meters give joules too. Mixing them costs a factor of a thousand.
5. Close the books with the first law, $Q = \Delta U - W$.

Checks. Work done on a gas being compressed is positive. For the same compression ratio, an adiabatic compression needs more work than an isothermal one, because it ends at a higher pressure and so its path encloses more area. The isothermal heat is always out of the gas during compression and into it during expansion. And a final temperature from $TV^{\gamma - 1}$ must be higher than the starting one after compression and lower after expansion.

7. Why the path matters

Take one mole of air at $300$ K and squeeze it to half its volume two different ways. Slowly, in a bath, the temperature stays at $300$ K and the work done on the gas is $8.31 \times 300 \times \ln 2 = 1730$ J; all of it flows out as heat, and the pressure doubles. Quickly, in an insulated cylinder, the temperature rises to $300 \times 2^{0.4} = 396$ K, the work done is $\tfrac{5}{2} \times 8.31 \times 96 = 1990$ J, no heat flows, and the pressure rises by $2^{1.4} = 2.64$.

Same start, same final volume, different work, different heat, different final state. Neither $W$ nor $Q$ is a property of the gas: each depends on how it got from one state to another. Their sum, $\Delta U$, does not — it depends only on the end points. That is why thermal physics distinguishes state functions such as $U$, $P$, $V$ and $T$ from the process quantities $W$ and $Q$, and why no gas can be said to "contain" a certain amount of heat.

8. In the world: why a diesel needs no spark

A gasoline engine compresses its fuel-air mixture about ten to one and fires it with a spark; a diesel compresses plain air about twenty to one and injects the fuel at the top of the stroke, where the air is already hot enough to ignite it. The adiabat says exactly how hot.

Air drawn in at $300$ K and squeezed to a twentieth of its volume in a few hundredths of a second has no time to lose heat to the cylinder walls. Its temperature rises by $20^{0.4} = 3.31$ to about $994$ K, or $720$ °C, far above diesel fuel's autoignition temperature of roughly $210$ °C. Its pressure rises by $20^{1.4} = 66$ to about $6.6$ MPa, which is why diesel engines are built heavier than gasoline ones. A gasoline engine at ten to one reaches only $300 \times 10^{0.4} = 754$ K; that is below gasoline's autoignition point in an engine, and a gasoline engine pushed to diesel ratios would ignite its fuel early, which is the knocking that high-octane fuel is designed to resist.

Real engines fall short of the ideal numbers because some heat does leak into the walls during the stroke, which is why a cold diesel can be hard to start and older diesels carried glow plugs to warm the cylinder first.

9. In the world: clouds from rising air

Air that rises through the atmosphere expands as the pressure around it falls, and because air conducts heat poorly, a large rising parcel exchanges almost no heat with its surroundings: it follows an adiabat. Along the way it cools.

Combining $PV^{\gamma}$ constant with the fall of pressure with height gives the dry adiabatic lapse rate, the cooling of rising dry air, of about $9.8$ K per kilometer. A parcel leaving the ground at $25$ °C is at about $15$ °C after one kilometer of ascent and $5$ °C after two. When it cools to its dew point, the water vapor it carries condenses into droplets, and a cloud forms; the flat bottoms of summer cumulus clouds mark the height where that happens. With a surface temperature of $25$ °C and a dew point of $13$ °C, the cloud base is roughly $(25 - 13)/8 \approx 1.5$ km up, using the forecaster's rule of about $8$ K per kilometer for the gap to close.

The same physics runs in reverse on the lee side of mountains: air descending the eastern slopes of the Rockies is compressed adiabatically and warms, producing the chinook winds that can raise the temperature in a Montana town by twenty degrees in an hour.

10. The sign, and the path

Two books can give opposite signs for the same work, and neither is wrong. Engineers often write the first law as $\Delta U = Q - W$ with $W$ the work done by the gas; physicists more often write $\Delta U = Q + W$ with $W$ the work done on it. This course uses the second throughout. The only safe habit is to say which you mean and check the sign physically: compressing a gas puts energy into it.

The deeper error is to think the work depends only on where the gas starts and ends. It is tempting because the end points are what the problem states. But the work is an area under a path, and different paths between the same two volumes enclose different areas. An isothermal squeeze and an adiabatic one between the same volumes do not even end in the same state. Only a state function — $U$, $T$, $P$, $V$ — has a change that ignores the route, and neither $W$ nor $Q$ is one.

11. Work at constant pressure

1. A gas at a steady $200$ kPa is compressed from $3.0$ L to $1.0$ L. Write the work done on it.

   $W = -P(V_f - V_i)$

   At constant pressure $P$ comes out of the integral $-\int P\,dV$.

2. Find the change in volume.

   $V_f - V_i = 1.0 - 3.0 = -2.0\ \text{L}$

   Compression gives a negative change in volume.

3. Convert to SI units.

   $P = 2.00 \times 10^{5}\ \text{Pa}, \qquad \Delta V = -2.0 \times 10^{-3}\ \text{m}^3$

   A litre is $10^{-3}$ m³.

4. Substitute and multiply.

   $W = -(2.00 \times 10^{5})(-2.0 \times 10^{-3}) = 400\ \text{J}$

   Two negatives make the work on the gas positive.

5. Check with the kilopascal-litre shortcut.

   $-200 \times (-2.0) = 400\ \text{kPa L} = 400\ \text{J}$

   $10^{3}\ \text{Pa} \times 10^{-3}\ \text{m}^3 = 1$ J, so kilopascals times litres are joules.

12. An isothermal compression

1. $0.10$ mol of gas in a bath at $300$ K is squeezed slowly from $2.0$ L to $1.0$ L. Write the pressure along the path.

   $P = \dfrac{nRT}{V}, \quad T = 300\ \text{K fixed}$

   The bath keeps the temperature constant while the volume changes slowly.

2. Set up the work integral.

   $W = -\int_{2.0}^{1.0}\dfrac{nRT}{V}\,dV = -nRT\left[\ln V\right]_{2.0}^{1.0}$

   $nRT$ is constant and comes outside.

3. Evaluate the logarithms.

   $W = -nRT(\ln 1.0 - \ln 2.0) = nRT\ln 2$

   $\ln a - \ln b = \ln(a/b)$, and the minus sign flips the ratio.

4. Work out $nRT$.

   $nRT = 0.10 \times 8.31 \times 300 = 249.3\ \text{J}$

   An energy, as it must be.

5. Multiply by the logarithm.

   $W = 249.3 \times 0.6931 \approx 173\ \text{J}$

   Positive: work done on the gas during compression.

6. Find the heat from the first law.

   $\Delta U = 0 \quad\Rightarrow\quad Q = -W \approx -173\ \text{J}$

   The temperature did not change, so the gas's thermal energy did not, and every joule of work left as heat into the bath.

13. The compression stroke of a diesel engine

1. Air at $300$ K and $100$ kPa is compressed adiabatically to $1/20$ of its volume. Find $\gamma$ for air.

   $f = 5, \qquad \gamma = \dfrac{f + 2}{f} = \dfrac{7}{5} = 1.4$

   Air is diatomic, with its vibrations frozen.

2. Write the temperature condition and solve for the final temperature.

   $T_fV_f^{0.4} = T_iV_i^{0.4} \quad\Rightarrow\quad T_f = T_i\left(\dfrac{V_i}{V_f}\right)^{0.4}$

   Along an adiabat $TV^{\gamma - 1}$ is fixed.

3. Substitute the compression ratio.

   $T_f = 300 \times 20^{0.4} = 300 \times 3.3145 \approx 994\ \text{K}$

   Hot enough to ignite diesel fuel sprayed in, which is why a diesel needs no spark plug.

4. Write the pressure condition and solve for the final pressure.

   $P_f = P_i\left(\dfrac{V_i}{V_f}\right)^{1.4}$

   Along an adiabat $PV^{\gamma}$ is fixed.

5. Substitute the compression ratio.

   $P_f = 100 \times 20^{1.4} = 100 \times 66.29 \approx 6.63 \times 10^{3}\ \text{kPa}$

   About $66$ atmospheres, three times what an isothermal squeeze would give.

6. Find the work done per mole from the first law.

   $Q = 0 \quad\Rightarrow\quad W = \Delta U = \tfrac{5}{2}R(T_f - T_i)$

   No heat flows, so the work is exactly the rise in thermal energy.

7. Substitute the temperatures.

   $W = 2.5 \times 8.31 \times (994 - 300) = 20.775 \times 694 \approx 1.44 \times 10^{4}\ \text{J per mole}$

   Positive, as compression work on the gas must be.

8. Compare with an isothermal compression of the same ratio.

   $W_{\text{iso}} = RT\ln 20 = 8.31 \times 300 \times 2.996 \approx 7.47 \times 10^{3}\ \text{J per mole}$

   The adiabat needs almost twice the work: it ends at a much higher pressure, so the area under its path is larger.

14. Your turn: helium at $300$ K is compressed adiabatically to half its volume. What is its temperature now?

1. Find the exponent for a monatomic gas.

   $\gamma - 1 = \dfrac{5}{3} - 1 = \dfrac{2}{3}$

   Helium has three degrees of freedom, so $\gamma = 5/3$.

2. Write the final temperature from the adiabatic condition.

   $T_f = T_i\left(\dfrac{V_i}{V_f}\right)^{2/3} = 300 \times 2^{2/3}$

   $TV^{\gamma - 1}$ is fixed.

3. Your turn: work this step out. Its working is at the end of the packet.

   Evaluate the power and multiply.

15. Guided practice

A gas held at a constant $500$ kPa expands from $8$ L to $16$ L. What is the work done on the gas?

• $-4000$ J
• $4000$ J
• $-8000$ J
• $0$ J, since the pressure does not change

16. Guided practice

Complete the worked solution: a monatomic ideal gas at a constant $100$ kPa is heated and expands by $9$ L. Find the work it does, the rise in its thermal energy, and the heat supplied.

1. Multiply the pressure by the change in volume.

   $P\Delta V = 100 \times 9 =$ _a J

   This is the work the gas does on its surroundings; a kilopascal times a litre is a joule.

2. Find the rise in thermal energy.

   $\Delta U = \tfrac{3}{2}nR\Delta T = \tfrac{3}{2}P\Delta V =$ _b J

   At constant pressure the ideal gas law gives $nR\Delta T = P\Delta V$.

3. Add the two to get the heat supplied.

   $Q = \Delta U - W = \Delta U + P\Delta V =$ _c J

   The work done on the gas is $W = -P\Delta V$, and the first law gives $Q = \Delta U - W$.

17. Guided practice

For $1$ mol of an ideal gas, match each process to what the first law $\Delta U = Q + W$ reduces to in it.

18. Practice

Helium starts at $300$ K and $100$ kPa and is taken along three different adiabats. Fill in its final temperature and pressure for each change of volume.

19. Practice

A cylinder of $0.7$ mol of an ideal gas sits in a water bath at $390$ K. A piston slowly squeezes the gas to $\tfrac{1}{10}$ of its volume. How much work is done on the gas? Use $R = 8.31$ J/(mol K).

Answer: unit: J / kJ / mJ

20. Practice

Air at $350$ K is compressed quickly, with no time for heat to escape, to $\tfrac{1}{20}$ of its volume. Treating air as a diatomic ideal gas with $\gamma = 1.4$, what is its temperature at the end?

Answer: unit: K

21. Somewhere new

A diesel engine draws in air at $100$ kPa and compresses it adiabatically to $\tfrac{1}{8}$ of its volume before fuel is injected. Treating air as an ideal gas with $\gamma = 1.4$, what is the pressure at the end of the compression?

Answer: unit: Pa / MPa / kPa

22. Lesson test

Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.

23. Test question

Helium starts at $400$ K and $200$ kPa and is taken along three different adiabats. Fill in its final temperature and pressure for each change of volume.

24. What you can do now

You can compute compression work and heat along the three standard paths. Explain to someone why squeezing a gas quickly takes more work than squeezing it slowly to the same volume.

Working for the steps left to you

14. Your turn: helium at $300$ K is compressed adiabatically to half its volume. What is its temperature now?, step 3