Theory — The First Law of Thermodynamics
Energy Conservation for a Gas
The First Law of Thermodynamics is the law of conservation of energy applied to a gas. The internal energy of the gas can change in two ways: by heat flowing in or out, and by work done on or by the gas as its volume changes.
ΔU = change in internal energy (J)
q = heat added to the gas (J)
W = work done on the gas (J)
This lab uses the convention that W is the work done on the gas. When a gas expands it does work on its surroundings, so the work done on the gas is negative (W = −PΔV); when it is compressed, work done on it is positive.
Internal Energy of an Ideal Gas
For an ideal gas, the internal energy depends only on temperature — not on pressure or volume separately. Raising the temperature raises the internal energy in direct proportion.
n = moles of gas, Cᵥ = molar heat capacity at constant volume
Monatomic ideal gas: Cᵥ = (3/2)R, Cₚ = (5/2)R, R = 8.314 J/mol·K
Work Done During a Volume Change
When the gas changes volume against a pressure, work is done. At constant pressure the work done on the gas is simply −P times the volume change.
Expansion (ΔV > 0): gas does work, W < 0
Compression (ΔV < 0): work done on gas, W > 0
Two Special Processes
Isochoric (constant V)
Volume fixed, so ΔV = 0 and W = 0. All the heat becomes internal energy: ΔU = q = n·Cᵥ·ΔT.
Isobaric (constant P)
Gas expands as it is heated. Heat splits: q = n·Cₚ·ΔT, work W = −PΔV, leaving ΔU = n·Cᵥ·ΔT.
Always
Whatever the process, ΔU = q + W. Internal energy is set by temperature alone.
| Process | Held constant | Work W (on gas) | First Law |
|---|---|---|---|
| Isochoric | Volume | 0 | ΔU = q |
| Isobaric | Pressure | −PΔV | ΔU = q − PΔV |
| Any ideal-gas process | — | varies | ΔU = q + W = n·Cᵥ·ΔT |
Apparatus
The equipment a real first-law-of-thermodynamics experiment uses. In the simulation these are modelled for you, but the readings correspond to what each instrument would measure.
Instructions — Running the Virtual Experiment
This is a measurement-and-calculation lab. For each scenario you will record the gas state and the energies from the simulation, then calculate the same quantities by hand from the First Law, and compare the two. Use n = 1.00 mol of a monatomic ideal gas, R = 8.314 J/mol·K, Cᵥ = 12.47 J/mol·K, and Cₚ = 20.79 J/mol·K. Record every value in your worksheet with units.
Simulation — Heating a Gas
Add heat (volume locked)
| T (K) | P (kPa) | V (L) | q (J) | W (J) | ΔU = q+W (J) |
|---|---|---|---|---|---|
| Add heat and click "Record point" to log each step. | |||||
Add heat (pressure constant)
| T (K) | P (kPa) | V (L) | q (J) | W (J) | ΔU = q+W (J) |
|---|---|---|---|---|---|
| Add heat and click "Record point" to log the process. | |||||
Team Questions
Example Lab Report
Sample report demonstrating the expected format and level of detail. Use as a guide for your own submission, and include labelled screenshots of the start and end states for each scenario.
The First Law of Thermodynamics
Physics | Section: [Your Section] | Date: [Date]
Lab Members: [Names of all members present]
Purpose
To verify the First Law of Thermodynamics, ΔU = q + W, for an ideal gas by carrying out two processes: an isochoric (constant-volume) heating, in which all the heat becomes internal energy, and an isobaric (constant-pressure) expansion, in which the heat divides between internal energy and the work done by the gas. The number of moles is found from the ideal gas law at the initial state.
Theory
The First Law states ΔU = q + W, where W is the work done on the gas. For an ideal gas the internal energy depends only on temperature, ΔU = n·Cᵥ·ΔT. At constant volume W = 0, so ΔU = q = n·Cᵥ·ΔT. At constant pressure the work done on the gas is W = −PΔV and the heat is q = n·Cₚ·ΔT, with Cₚ = Cᵥ + R. A monatomic ideal gas has Cᵥ = (3/2)R = 12.47 J/mol·K and Cₚ = (5/2)R = 20.79 J/mol·K.
Isochoric: W = 0, ΔU = q · Isobaric: W = −PΔV, q = n·Cₚ·ΔT
Determining the Moles (initial state)
n = PV/RT = (100000 Pa)(0.0249 m³)/[(8.314)(300)] = 1.00 mol
Scenario 1 — Isochoric (V = 24.9 L constant)
Measured: T: 300 → 400 K, P: 100.0 → 133.3 kPa, V constant
Work: W = −PΔV = 0 (no volume change)
Internal energy: ΔU = n·Cᵥ·ΔT = (1.00)(12.47)(100) = 1247 J
Heat: q = ΔU = 1247 J, confirming ΔU = q when W = 0.
Scenario 2 — Isobaric Expansion (P = 100 kPa constant)
| T (K) | P (kPa) | V (L) | q (J) | W (J) | ΔU = q+W (J) |
|---|---|---|---|---|---|
| 300 | 100 | 24.9 | 0 | 0 | 0 |
| 325 | 100 | 27.0 | 520 | −208 | 312 |
| 350 | 100 | 29.1 | 1039 | −416 | 624 |
| 375 | 100 | 31.2 | 1559 | −624 | 935 |
| 400 | 100 | 33.3 | 2079 | −833 | 1247 |
At each step q + W equalled n·Cᵥ·ΔT, and the final ΔU of 1247 J matched the isochoric result for the same temperature change.
Discussion
In the isochoric process the volume did not change, so no work was done and all 1247 J of heat went into internal energy — the readout confirmed ΔU = q. In the isobaric expansion, 2079 J of heat was added but the gas did 833 J of work pushing the piston out (W = −833 J on the gas), leaving 1247 J as the internal-energy increase. Both processes raised the temperature by the same 100 K, and indeed both gave the same ΔU = 1247 J, illustrating that for an ideal gas the internal energy depends only on temperature, not on the path taken.
The numerical check q + W = ΔU held at every step of the expansion, confirming the First Law. Small differences between calculated and on-screen values came from reading precision and from averaging the pressure while the piston moved. The constant-volume case had the lowest uncertainty because only temperature and pressure had to be read.
Conclusion
The First Law of Thermodynamics was confirmed: in constant-volume heating ΔU ≈ q (W = 0), while the expansion required the work term, with q + W = ΔU throughout. Internal energy was set by temperature alone, and the results were consistent with ideal-gas thermodynamics within reading precision.
Practice Questions
Show all work and include units. Use R = 8.314 J/mol·K; for a monatomic ideal gas Cᵥ = 12.47 and Cₚ = 20.79 J/mol·K.