Unit 2 · Lab

Boyle’s law and work from a PV curve

Measure how the pressure of a trapped gas depends on its volume, linearize the result to prove it is a genuine inverse relationship, then integrate your own data to find the work done compressing it. One afternoon, one syringe, and every idea in the unit shows up.

~60 minutes · equipment: minimal

Research question
Theory
Materials
Safety
Procedure
Data table
Analysis
Sample data & worked analysis
Error analysis
Conclusion questions
Extensions

Research question

For a fixed quantity of gas held at constant temperature, how does absolute pressure depend on volume — and how much work is required to compress it?

Hypothesis to test: P ∝ 1/V, so a graph of P against 1/V should be a straight line through the origin with slope nRT.

Theory

Starting from PV = nRT, hold n and T fixed and the right-hand side is a constant:

PV = constant ⇒ P = (nRT) · 1V Plotting P vs 1/V turns a hyperbola into a line — and a straight line is something your eye can actually check.

A curved graph is weak evidence: many different relationships produce curves that look alike. Linearizing is the whole method. If P vs 1/V comes out straight, the inverse relationship is confirmed; if it bends, it is not.

The work done on the gas is the area under the PV curve:

W_on = −∫_{V_i}^V_f P dV and for an isothermal process W_on = −nRT ln(V_fV_i)

You will compute this two ways — numerically from your measured points, and analytically from the formula — and see whether they agree. That comparison is the real payoff of the lab.

Materials

Two versions. Do whichever your equipment allows; the analysis is identical.

Version A — with a pressure sensor (preferred)

60 mL plastic syringe with a Luer fitting
Gas pressure sensor (Vernier, PASCO, or equivalent) with short tubing
Data logger, tablet, or laptop
Thermometer for ambient temperature
Ruler or the syringe’s own graduations

Version B — no sensor, using known masses

Syringe mounted vertically, nozzle sealed with a cap or plug
Set of slotted masses (0.5–5 kg total) and a platform that rests on the plunger
Calipers to measure the plunger diameter (you need the area A)
Vacuum grease or a drop of oil to reduce plunger friction

In version B, pressure is not measured — it is computed from the force balance on the plunger:

P = P_atm + mgA A = πd²/4 for the plunger. Look up the day’s local barometric pressure — do not assume exactly 101.3 kPa.

Version A. The tubing and sensor cavity hold gas too — that extra “dead volume” is the main systematic error in this experiment, and part 4 of the analysis measures it.

Safety

Do not compress below about 1/4 of the starting volume. Pressure climbs steeply and a plastic syringe can crack or a fitting can blow off.
Keep a hand on the plunger at all times when compressed — a released plunger fires outward.
Never point the nozzle at anyone, and wear eye protection while compressing.
In version B, load masses gently and centered; an off-center stack tips.

Procedure

Record the ambient temperature and, if you have a barometer or a weather app, the local atmospheric pressure. Convert the temperature to kelvin immediately so it is not forgotten later.
Draw the plunger to 30.0 mL with the syringe open, then connect it to the sensor (version A) or cap it (version B). The gas is now sealed.
With the plunger at rest, record the starting pressure. It should be close to atmospheric — if it is not, your seal is leaking. Fix that before continuing.
Compress to the next volume on your list. Hold it steady and wait 10–15 seconds before recording the pressure. See the callout below — this step is the difference between a good data set and a bad one.
Record pressure and volume. Repeat for volumes of 25, 20, 17.5, 15, 12.5, and 10 mL.
Now reverse: let the plunger back out through the same volumes, recording pressure again. Comparing the two sweeps reveals friction and leaks.
Repeat the entire run three times and average. If run 3 reads systematically lower than run 1 at every volume, you have a slow leak.

The step everyone skips

Compressing a gas quickly is adiabatic, not isothermal — the gas heats up and reads high, exactly as described on the first law page. Waiting 10–15 seconds lets that extra energy leak out through the syringe walls so the gas returns to room temperature. Skip the wait and your points will sit systematically above the true isotherm, and the run will not linearize cleanly.

That failure mode is itself a nice observation: watch the pressure reading drift downward for a few seconds after each compression. You are watching a process go from adiabatic to isothermal in real time.

Data table

Print this page or copy the table. Absolute pressure, not gauge.

Ambient temperature: ________ °C = ________ K Atmospheric pressure: ________ kPa

The last column is a running sanity check: if Boyle’s law holds, every entry should be the same number to within a few percent.
`V` (mL)	1/`V` (mL⁻¹)	`P` run 1 (kPa)	`P` run 2 (kPa)	`P` run 3 (kPa)	`P` avg (kPa)	`PV` (kPa·mL)
30.0	0.0333
25.0	0.0400
20.0	0.0500
17.5	0.0571
15.0	0.0667
12.5	0.0800
10.0	0.1000

Analysis

1. The raw PV curve

Plot P (vertical) against V (horizontal). You should get a hyperbola falling to the right — the isotherm from the graph gallery, this time made of your own measurements.

2. Linearize

Plot P against 1/V and fit a straight line. Report the slope with units and the vertical intercept.

Straight line ⇒ P ∝ 1/V confirmed.
Slope = nRT, in kPa·mL if you kept those units.
Intercept should be zero. A noticeably positive one is a real signal, not noise — see part 4.

3. Extract the amount of gas

Convert the slope to SI (1 kPa·mL = 10⁻³ J) and solve:

n = slopeRT Then N = nN_A gives the number of molecules, and m = nM with M_air ≈ 0.029 kg/mol gives the mass of air you trapped.

Check it against the direct calculation n = PV/RT using your starting state. The two should agree closely — if they do not, suspect the temperature or a unit conversion.

4. Find the dead volume

The syringe graduations do not include the gas sitting in the nozzle, tubing, and sensor cavity. Calling that extra volume V_d, the true relation is

P(V_read + V_d) = nRT

which is why a plot against 1/V_read shows a small positive intercept and a slight bend. To measure V_d, try adding a trial value to every volume and re-plotting; the value that makes the line straightest and drives the intercept to zero is your dead volume. A few tenths of a millilitre up to a couple of millilitres is typical.

5. Work from the area — numerically

Use the trapezoidal rule on your own P–V points. For consecutive points,

W_by ≈ Σ P_i + P_i+12 (V_i+1 − V_i) Each term is the area of one trapezoid under the curve. Watch your units: kPa·mL = 10⁻³ J.

Compute the total for the compression from 30.0 mL down to 10.0 mL. Since the gas is being compressed, W_on is positive and equals that area.

Calculus corner

Because the curve is convex (it bends upward), the trapezoidal rule overestimates the true area — each straight chord sits above the curve. With seven points the error is small, but it is real and it is systematic, so it belongs in your error discussion rather than being hidden in “random uncertainty.” Taking more, closer-spaced points at the high-pressure end, where the curve bends most, shrinks it fastest.

6. Work from the formula — analytically

Compare your numerical area to the isothermal prediction, using the nRT you got from the slope:

W_on = −nRT ln(V_fV_i) With V_f < V_i the logarithm is negative, so W_on comes out positive. Include the dead volume in both volumes if you measured it.

Report the percent difference between the two methods.

Sample data & worked analysis

Use this to practice the analysis before you have your own data — or to check your spreadsheet is doing what you think. Taken at 22 °C (295 K).

The `PV` column drifts down by about 6% from top to bottom — not random scatter, but a systematic trend. That drift is the dead volume announcing itself.
`V` (mL)	1/`V` (mL⁻¹)	`P` (kPa)	`PV` (kPa·mL)
30.0	0.0333	68.5	2055
25.0	0.0400	81.7	2043
20.0	0.0500	101.6	2032
17.5	0.0571	114.5	2004
15.0	0.0667	133.1	1997
12.5	0.0800	156.2	1953
10.0	0.1000	192.4	1924

Straight to the eye, which confirms P ∝ 1/V. The intercept is about +7 kPa rather than 0 — small, but consistently present, and exactly what a dead volume of roughly 1 mL produces.

Worked analysis of the sample data

Slope. Using the first and last points:

slope = 192.4 − 68.50.1000 − 0.0333 = 123.90.0667 = 1857 kPa·mL

Converting: 1857 kPa·mL = 1.857 J. (Two-point slopes are crude — use a least-squares fit on the real thing.)

Amount of gas.

n = 1.857(8.31)(295) = 7.6×10⁻⁴ mol ≈ 4.6×10²⁰ molecules, or about 22 mg of air.

Dead volume. The data are fit well by

P(V + 1.2 mL) = 2148 kPa·mL Check at V = 10.0: 2148/11.2 = 191.8 kPa, against 192.4 measured. At V = 30.0: 2148/31.2 = 68.8, against 68.5. Both within 0.5%.

With the dead volume included the corrected slope is 2148 kPa·mL, giving n = 8.8×10⁻⁴ mol — about 15% higher than the uncorrected value. Systematic errors matter.

Work, numerically. Trapezoids from 10.0 to 30.0 mL:

10.0→12.5:  ½(192.4+156.2)(2.5) =  435.8
12.5→15.0:  ½(156.2+133.1)(2.5) =  361.6
15.0→17.5:  ½(133.1+114.5)(2.5) =  309.5
17.5→20.0:  ½(114.5+101.6)(2.5) =  270.1
20.0→25.0:  ½(101.6+ 81.7)(5.0) =  458.3
25.0→30.0:  ½( 81.7+ 68.5)(5.0) =  375.5
                            total  = 2210.8 kPa·mL = 2.21 J

So compressing from 30.0 mL to 10.0 mL requires W_on = +2.21 J.

Work, analytically. Using the dead-volume-corrected constant:

W_on = −(2148) ln(11.231.2) = (2148)(1.0245) = 2201 kPa·mL = 2.20 J

Percent difference: 0.5%. The numerical result is slightly higher, exactly as the convexity argument predicts. Two independent methods agreeing to half a percent is a strong result — and the direction of the small disagreement is itself explained by the mathematics, which is better than agreement alone.

For scale: 2.2 J is roughly the work of lifting a 250 g apple one metre. You can feel it in your hand as you push the plunger.

Error analysis

Separate the two kinds — AP rubrics award credit for knowing the difference.

Source	Type	Effect & fix
Dead volume in tubing and nozzle	Systematic	Every measured `V` is too small, so `PV` drifts. Fix by fitting `V`_d as in part 4, or by using the shortest possible tubing.
Compressing too fast (adiabatic heating)	Systematic	Pressures read high; the curve is steeper than the true isotherm. Fix by waiting for equilibrium at every point.
Plunger friction	Systematic, direction-dependent	Makes compression readings high and expansion readings low — which is why the two sweeps disagree. Average them, or lubricate the plunger.
Slow leak past the seal	Systematic, grows with time	Later runs read low at every volume. Detect by comparing run 1 with run 3; fix the seal rather than averaging it away.
Reading the graduations	Random	±0.25 mL is typical, so it is worst at small volumes — 2.5% at 10 mL versus 0.8% at 30 mL. Repeat and average.
Room temperature drift	Random / systematic	`nRT` is not quite constant across a long session. Record `T` at the start and end.
Trapezoidal approximation	Systematic	Overestimates the area for a convex curve. Shrink it with more points where the curvature is greatest.

What makes this an AP-style analysis

Notice that every systematic error above has a known sign. “Human error” earns nothing on a rubric; “dead volume makes every recorded volume too small, which lowers the apparent PV product most at small volumes” earns full credit. Always state the direction of an error, not just its existence.

Conclusion questions

Does your P vs 1/V graph support the hypothesis? Cite the shape and the intercept, not just “it looked straight.”
Your PV column probably drifts in one direction. Which direction, and which systematic error explains that specific direction?
The gas got warmer as you compressed it, yet you analyzed the data as isothermal. Justify that, referring to what you did in the procedure.
Using the first law, state the sign of Q for the compression. (You did positive work on the gas and its temperature ended unchanged — so where did the energy go?)
Sketch your compression on a PV diagram and shade the region whose area you computed. Which page’s gallery figure does it match?
If you repeated the lab with the syringe in an ice bath at 0 °C, how would the slope of your linearized graph change? By what factor, quantitatively?
Estimate how much your result would change if the dead volume were twice what you measured. Is this experiment more sensitive to the dead volume or to your reading uncertainty?

Extensions

Catch the adiabat. Compress the syringe as fast as you can while logging at a high sample rate, then hold. Plot the fast compression and the settled points on the same axes. The fast curve is steeper — you have measured an adiabatic process and an isothermal one on the same apparatus, which is precisely the comparison figure from the notes.
Charles’s law. Keep the plunger free to move (constant pressure) and put the syringe in water baths at several temperatures. Plot V vs T in kelvin; extrapolate the line to V = 0 and you get an estimate of absolute zero.
Find k_B. Combine your n from the slope with Avogadro’s number to get k_B = R/N_A, and compare to the accepted value.
Round trip. Compress, then expand back along a different route by changing the temperature partway. Compute the enclosed area — you have built a (very inefficient) heat engine, and its net work is the loop area from the second law page.