Interactive Laboratories

Ideal gas & PV work

The box on the left holds a fixed amount of monatomic gas (n = 0.10 mol of argon). Move the piston or change the temperature and watch the state point travel across the PV diagram on the right. Pick a process first and the simulation will hold the right quantity constant for you.

Thermal conductivity

A slab bridges a hot and a cold reservoir. The steady-state transfer rate is P = kAΔT/L — proportional to conductivity, area and temperature difference, and inversely proportional to thickness.

Entropy

Move a fixed amount of heat between two reservoirs and watch the entropy budget. The hot side loses Q/T_H, the cold side gains Q/T_C, and because T_C is the smaller denominator the total comes out positive — only in the hot → cold direction.

Heat transfer & calorimetry

Drop two samples into an insulated container. No energy leaves, so ΣQ = 0 and the mixture settles wherever the two mcΔT terms cancel.

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.

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:

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:

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:

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

1. 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.
2. 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.
3. 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.
4. 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.
5. Record pressure and volume. Repeat for volumes of 25, 20, 17.5, 15, 12.5, and 10 mL.
6. Now reverse: let the plunger back out through the same volumes, recording pressure again. Comparing the two sweeps reveals friction and leaks.
7. 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.

Data table

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

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

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:

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

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,

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.

6. Work from the formula — analytically

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

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).

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

Error analysis

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

Conclusion questions

1. Does your P vs 1/V graph support the hypothesis? Cite the shape and the intercept, not just “it looked straight.”
2. Your PV column probably drifts in one direction. Which direction, and which systematic error explains that specific direction?
3. 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.
4. 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?)
5. Sketch your compression on a PV diagram and shade the region whose area you computed. Which page’s gallery figure does it match?
6. 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?
7. 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.

Theory for this lab: Unit 1 · PV diagrams →

Coulomb force & superposition

Drag the point charges around and set their sizes with the sliders. The simulation draws every individual Coulomb force on the charge you pick (dashed) and their vector sum (solid). Each pair obeys F = k|q₁q₂|/r² on its own; the other charges change nothing about it. The net force is just the vectors added tip to tail.

Theory for this lab: Unit 2 · Coulomb’s law →

Field lines & equipotentials

The electric field and the electric potential are two pictures of the same charges. Field lines start on positive charge and end on negative charge; equipotential lines join points at the same V = Σkq/r. Wherever they meet they cross at right angles, and where the equipotentials bunch together the field is strong, because |E| ≈ |ΔV/Δs|.

Theory for this lab: Unit 2 · Electric potential →

Charged particle between plates

Two parallel plates with a potential difference ΔV make a uniform field E = ΔV/d between them. A particle that enters horizontally feels a constant force qE straight across the gap and none along it, so it follows a parabola — exactly like a projectile, with qE/m in place of g.

Theory for this lab: Unit 2 · Electric field →

Parallel-plate capacitor

Geometry alone sets the capacitance: C = κε₀A/d. What happens to Q, ΔV, E and the stored energy when you change that geometry depends on one question: is the battery still connected? Connected, the battery holds ΔV fixed. Disconnected, the charge has nowhere to go, so Q is fixed.

Theory for this lab: Unit 2 · Capacitors →

Mapping equipotentials and field lines

You cannot see an electric field, but you can measure potential with an ordinary voltmeter. Paint two electrodes onto conductive paper, hunt for points that read the same voltage, and the invisible field draws itself: equipotentials first, field lines at right angles to them, and the field strength from how closely the lines are packed.

Research question

What do the equipotentials and field lines look like between two charged electrodes, and does the field between two parallel bars have the uniform magnitude |E| = ΔV/d?

Hypothesis to test: between long parallel electrodes the equipotentials are straight, parallel and evenly spaced, so the potential rises linearly with distance from the negative electrode, with slope equal to the field magnitude.

Theory

An equipotential is a line (in 3-D, a surface) on which every point has the same electric potential. Moving a charge along one takes no work, since W = −qΔV = 0. That forces the field to have no component along the equipotential, so:

• Field lines cross equipotentials at right angles.
• The field points from high potential toward low potential.
• Crowded equipotentials mean a strong field, because the same ΔV is dropped over a shorter distance.

The magnitude of the field is the rate at which the potential changes with distance:

Why conductive paper works. The paper is a weak conductor, so the supply drives a small steady current through it. Inside the paper that current flows along the field lines, and the potential obeys the same equation it would in empty space around the same electrodes (in two dimensions). The paper gives the voltmeter something to touch: a map of V on the paper is a map of the field’s potential.

Materials

• Conductive (carbon-loaded) paper, about 20 cm × 28 cm — ideally with a printed 1 cm grid
• Silver conductive-ink pen for drawing electrodes
• Corkboard or foam board, and metal push pins
• DC power supply, 6–12 V (or a 9 V battery)
• Digital multimeter on the DC volts setting, with a fine-tipped probe
• Connecting leads with alligator clips
• Ruler, and a sheet of ordinary grid paper for recording the map

Safety

• The voltages are low, but never connect the supply leads directly to each other — a short can overheat leads and damage the supply. Always connect through the electrodes.
• Let the silver ink dry completely (at least 15–20 minutes) before powering. Work in a ventilated room and keep the ink off skin.
• Push pins are sharp; press them in with a thumb guard or the back of a ruler.
• Turn the supply off before rearranging any connection.

Procedure

1. Draw two straight, parallel bar electrodes 12.0 cm long and 10.0 cm apart in the middle of the conductive paper. Make each line thick, unbroken and evenly inked — gaps and thin patches become extra resistance. Let it dry.
2. Pin the paper to the board. Push a metal pin through each electrode and clip a lead to each pin. Connect the leads to the supply and set it to about 10 V.
3. Clip the voltmeter’s COM lead to the negative electrode. Touch the probe to the positive electrode and record the reading as ΔV_supply. Touch it to the negative electrode: it should read 0.00 V.
4. Center line. Along the line midway between the bar ends (y = 0), record V every 1.0 cm from x = 0 (on the 0 V bar) to x = 10.0 cm (on the positive bar).
5. Equipotentials. For each target voltage — 2.00, 4.00, 6.00 and 8.00 V — slide the probe along the lines y = −3.0, 0 and +3.0 cm until the meter reads the target. Record each x. Then find a few more points of each equipotential beyond the ends of the bars (for example y = ±7 cm and ±9 cm) to see how the lines curve.
6. Transfer every point to grid paper, with the electrodes drawn in. Join points of equal voltage with a smooth curve and label it.
7. Field lines. Starting on the positive electrode, draw lines that cross every equipotential at a right angle and end on the negative electrode. Add arrows pointing from high to low potential.

Data table

Supply voltage across electrodes: ________ V    Electrode separation d: ________ cm

Analysis

1. The map

Look at your finished drawing. Between the bars, are the equipotentials straight, parallel to the electrodes, and evenly spaced? Where do they start to curve, and which way? Do your field lines meet both electrodes at right angles? (They should — each electrode is itself an equipotential.)

2. Potential vs. position

Plot Table 1 as V (vertical) against x (horizontal). Fit a straight line to the points between the electrodes, leaving out the two points measured on the silver itself. A straight line means the field is uniform.

3. Compare with the parallel-plate prediction

Compute ΔV_supply/d and the percent difference from your slope. Decide whether any gap is random scatter or a systematic effect — and if systematic, which way it points.

4. Field from equipotential spacing

The same field can be read straight off the map. Adjacent equipotentials in Table 2 differ by 2.00 V, so for each gap:

Sample data & worked analysis

Use this to practice before you have your own data. Supply set to 10.0 V; electrodes 10.0 cm apart, 12.0 cm long.

2.00→4.00 V:  2.00 V / 0.0216 m = 92.6 V/m
4.00→6.00 V:  2.00 V / 0.0218 m = 91.7 V/m
6.00→8.00 V:  2.00 V / 0.0217 m = 92.2 V/m
                          mean = 92.2 V/m

Error analysis

Separate the two kinds, and give each systematic error a direction.

Conclusion questions

1. Explain, using the work done moving a charge along an equipotential, why field lines must cross equipotentials at right angles.
2. Where on your map is the field strongest, and where weakest? How can you tell from the equipotentials alone, without drawing a single field line?
3. Does your V vs x graph support a uniform field? Cite the shape, the residuals, and the equipotential spacings.
4. Your slope and ΔV_supply/d probably disagree. Which is larger, and which systematic error explains that specific direction?
5. If this were a vacuum gap with the same potentials, how much work would the field do on a proton moving from the 8.00 V equipotential to the 2.00 V equipotential? Give the answer in joules and in electron-volts. Would it be different for an electron, and how?
6. Why is a voltmeter (and not an ammeter) the right instrument for this mapping, and why must it draw very little current?
7. Predict how your map would change if the supply were raised to 12 V. Which quantities change, and which stay the same?

Extensions

• A “dipole.” On a fresh sheet, paint two small filled circles 10 cm apart as point-like electrodes. Map the 2, 4, 5, 6 and 8 V equipotentials and compare with the field-line pattern of two opposite charges on the electric field page. The 5 V line should be the perpendicular bisector — and the equipotentials should crowd near each electrode, where the field is strongest.
• Shielding. Paint a closed ring electrode with a small dot electrode outside it, and connect the supply between them. Probe inside the ring: every point reads the same voltage as the ring, so the field inside is zero — the conductor-in-equilibrium result from the conductors & capacitors page.
• Field vs. voltage. Repeat the center-line measurement at 4, 6, 8, 10 and 12 V. Plot the slope against the supply voltage: it should be a straight line through the origin, |E| ∝ ΔV for fixed d.
• Compare with the simulation. Set up the same electrode geometry in the field & equipotential map and lay your hand-drawn map beside it.

Theory for this lab: Unit 2 · Electric potential →

Ohm’s law & resistivity

A uniform wire is connected across a fixed potential difference. Its resistance comes only from what it is made of and its shape, R = ρL/A; the current then follows from I = ΔV/R. Change the material, the length or the thickness and watch the I–V line on the right tilt.

Theory for this lab: Unit 3 · Current & resistance →

Series vs parallel bulbs

Three bulbs, one ideal battery, four ways to wire them. Brightness tracks the power each bulb dissipates, P = IΔV. Pull a bulb out of its socket and see which of the others go dark, which get dimmer and which get brighter.

Theory for this lab: Unit 3 · Series & parallel →

Two-loop circuit solver

Two batteries, three resistors, two junctions: the circuit that cannot be reduced to series and parallel pieces. The simulation solves it exactly and then checks its own answer with Kirchhoff’s junction rule and both loop rules, using the actual numbers.

Theory for this lab: Unit 3 · Kirchhoff’s rules →

RC charging & discharging

Throw the switch to connect the capacitor to the battery through a resistor, then throw it the other way to let the capacitor drain through the same resistor. Everything happens on the time scale τ = RC: after one τ the capacitor has gone 63% of the way to where it is heading.

Theory for this lab: Unit 3 · RC circuits →

Ohm’s law and the resistivity of a wire

Stretch a length of nichrome wire along a metre rule, push a current through it, and measure the voltage across different lengths. First show the wire obeys Ohm’s law, then turn the slope of a straight-line graph into the resistivity of the alloy — a material constant you can check against a table.

Research question

Is a nichrome wire ohmic, and what is the resistivity of nichrome?

Hypotheses to test: (A) at fixed length, ΔV ∝ I, so a graph of ΔV against I is a straight line through the origin; (B) at fixed current, R ∝ L, so a graph of R against L is a straight line whose slope is ρ/A.

Theory

For any conductor, resistance is defined by

For a uniform wire, resistance depends on its geometry and its material:

Using the slope rather than a single R/L value is the point of the method: any constant offset — a contact that sits a few millimetres off the zero mark, for instance — shifts the line up or down but leaves the slope alone.

Nichrome is chosen deliberately. Its resistivity is high (so a metre of thin wire has a measurable resistance of order 10 Ω) and nearly independent of temperature (so the wire stays ohmic even as it warms). Try the same experiment with copper and the resistance is too small to measure well; try it with a bulb filament and it is not ohmic at all.

Materials

• About 1.1 m of bare nichrome wire, roughly 0.3 mm diameter (around 28–30 gauge)
• Metre rule, with the wire taped or clamped taut along it
• Variable DC power supply, 0–12 V (or a battery pack plus a rheostat)
• Ammeter (0–1 A range) and voltmeter (0–5 V range), or two digital multimeters
• Switch, connecting leads, two crocodile clips, and a sliding contact (“jockey”)
• Micrometer screw gauge

Safety

• Keep the current at or below 0.5 A. The wire dissipates a few watts and thin nichrome gets hot enough to burn skin above that.
• Open the switch between readings. This also keeps the wire near room temperature, which keeps its resistance steady.
• Never connect the ammeter across the supply or across the wire — it has almost no resistance and is a short circuit.
• Low-voltage DC only. Do not substitute a mains-powered device for the supply.

Procedure

1. Measure the diameter with the micrometer at three points along the wire, and at each point in two perpendicular directions (the wire may not be perfectly round). Check the micrometer’s zero reading first and correct for it. Record all six readings.
2. Tape the wire taut along the metre rule, avoiding kinks. Build the circuit in the figure: supply, switch and ammeter in series with the full wire; voltmeter clip fixed at the 0 cm mark.
3. Part A. Place the jockey at L = 50.0 cm. Set the supply so the current is 0.100 A, close the switch, and record I and ΔV. Open the switch.
4. Repeat for currents of 0.200, 0.300, 0.400 and 0.500 A, opening the switch between readings.
5. Part B. Set the current to 0.300 A. Record ΔV with the jockey at L = 20.0, 40.0, 60.0, 80.0 and 100.0 cm. Check that I stays at 0.300 A as you go — it should, since the circuit itself is not changing.
6. Press the jockey on lightly and read the voltmeter promptly; dragging it scrapes the wire and changes its cross-section.
7. Repeat Part B once more, moving the jockey in the opposite order, and average.

Data table

Wire diameter readings (mm): ______ ______ ______ ______ ______ ______    Mean d: ______ mm    Zero error: ______ mm

Part A — fixed L = 50.0 cm

Part B — fixed I = 0.300 A

Analysis

1. Is the wire ohmic?

Plot ΔV (vertical) against I (horizontal) for Part A and fit a straight line. A straight line through the origin confirms ohmic behaviour; its slope is the resistance of 50.0 cm of wire. (With the axes this way round the slope is R; plotted as I vs ΔV it would be 1/R.)

2. Resistance against length

For Part B compute R = ΔV/I at each length, plot R against L, and fit a straight line. Report the slope in Ω/m and the intercept in Ω.

3. Cross-sectional area

4. Resistivity

As a cross-check, compute ρ = RA/L from Part A alone, using the slope of the ΔV–I graph and L = 0.500 m. The two methods should agree.

Sample data & worked analysis

Use this to practice the analysis before you have your own data, or to check your spreadsheet. Micrometer readings (mm): 0.321, 0.318, 0.322, 0.319, 0.320, 0.320; zero error 0.000 mm. Mean d = 0.320 mm.

Error analysis

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

Conclusion questions

1. Does your Part A graph show the wire is ohmic? Cite the shape and the intercept, not just “it looked straight.”
2. Explain why moving the jockey does not change the ammeter reading.
3. Your R–L graph probably has a small nonzero intercept. Give one physical cause and explain why it does not affect your value of ρ.
4. Which single measurement contributes most to the uncertainty in ρ? Justify with percentages.
5. Predict the slope of the R–L graph for a nichrome wire of twice the diameter. By what factor does it change?
6. If the same experiment were done with a copper wire of the same diameter, what slope would you expect, and why would it be hard to measure with this equipment?
7. Sketch the ΔV–I graph you would get if the wire were replaced with a small filament bulb, and explain its shape using the temperature dependence of resistance.

Extensions

• Test the 1/A dependence. Repeat Part B with two or three other gauges of nichrome. Plot the slope against 1/A; it should be a straight line through the origin with slope ρ.
• Non-ohmic contrast. Replace the wire with a 6 V filament bulb and take ΔV against I from 0 to 6 V. Compute R = ΔV/I at each point and plot it against power to see the filament heating up.
• Temperature coefficient. Coil a length of copper wire in a water bath and measure its resistance from 20 °C to 80 °C. The rise is a few tenths of a percent per degree — a direct look at more lattice collisions.
• Power in the wire. For each Part A reading, compute P = IΔV and confirm it equals I²R — connecting this lab to electric power.

Theory for this lab: Unit 3 · Current & resistance →