Preview

Ideal Gas Law & PV Diagrams

Visualize isothermal, isobaric, isochoric, and adiabatic processes

Seal two moles of nitrogen in a five-liter box and nothing seems to happen. Nothing visible, anyway — inside, a trillion trillion molecules are hammering the walls billions of times per second, and that drumbeat has a name: pressure. This lab hands you the microscopic engine room of the macroscopic gas laws. The P row is not decoration — it is bookkeeping, P = nRT/V, every pascal earned by molecular collisions. Heat the gas and the molecules' average kinetic energy climbs in lockstep: K̄ = (3/2)k_B·T, a rule so strict it ignores what the molecules are made of — helium and xenon at the same temperature carry the same average energy; the light ones simply move faster to make up for it. Check the v_rms readout: at 300 K, nitrogen molecules cross the box at 517 m/s — faster than sound in air — yet the gas looks perfectly still, because the chaos is isotropic. Then open the Maxwell-Boltzmann board: not every molecule got the memo. Speeds spread over a wide, single-peaked distribution whose area is nailed to 1 forever — heat the gas and the curve slides right and flattens, but it can never just grow taller. Three sliders — temperature, volume, moles — one equation, one speed distribution, and every gas law your textbook named after a dead Frenchman falls out. Learn this state bookkeeping well, because the sequel spends it: a heat engine is just this same gas, pushed around a loop.

What you'll be able to do

  • Apply PV = nRT with the panel's own numbers: at the default state (n=2.0 mol, V=5.0 L, T=300 K) the P row reads 997.7 kPa; seal the box and heat to 600 K and it doubles exactly to 1995.4 kPa — 'at constant volume and number of molecules, the pressure is proportional to the temperature' (OpenStax §2.1, Gay-Lussac's law); squeeze the volume 10→1 L at 300 K and P climbs 498.8→4988.4 kPa, inversely (Boyle); add gas 0.5→5.0 mol and P tracks it linearly, 249.4→2494.2 kPa
  • Connect temperature to microscopic motion: the Avg KE/molecule row is K̄ = (3/2)k_B·T — 'nothing in this equation depends on the molecular mass (or any other property) of the gas… helium and xenon… at the same temperature… have the same average kinetic energy' (§2.2 verbatim) — so at 600 K the row reads 12.43×10⁻²¹ J for ANY gas, while v_rms = √(3RT/M) rises only by √2 (517→731 m/s for nitrogen) because the square lives inside the speed, not the energy
  • Read the Maxwell-Boltzmann board as a probability distribution: its area is pinned to 1 ('∫₀^∞ f(v)dv = 1', §2.4), so as T climbs 300→1000 K the distribution 'is shifted to higher speeds and broadened at higher temperatures' (Fig. 2.16 caption) — wider AND shorter, never taller in place — while the cyan v_rms marker marches 517→944 m/s across the board

Formulas

PV=nRT(R=8.314Jmol1K1)PV = nRT \quad (R = 8.314\,\text{J}\cdot\text{mol}^{-1}\text{K}^{-1})
Ideal Gas Law
W=PΔVW = P\Delta V
Work done in isobaric process
W=nRTln ⁣(V2V1)W = nRT\ln\!\left(\frac{V_2}{V_1}\right)
Work done in isothermal process
ΔU=QW\Delta U = Q - W
First Law of Thermodynamics

Make a prediction

Default state: 2.0 mol of gas in a 5.0 L box at 300 K — the panel reads P = 997.7 kPa. You seal the box (n and V locked) and drag Temperature from 300 K to 600 K. What does the P row read?

No grading here — commit to a guess, then scroll down and test it yourself.

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Check

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Your prediction

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Answer: 1995.4 kPa — exactly double: at constant V and n, P = (nR/V)·T is proportional to ABSOLUTE temperature, and 600 K is twice 300 K

Gay-Lussac's law, stated flat in the textbook: 'at constant volume and number of molecules, the pressure is proportional to the temperature' (OpenStax §2.1). The mechanism lives one section over — pressure is the averaged momentum transfer of wall collisions, and 'if the average velocity of the molecules is higher, the gas pressure is higher' (§2.2). Watch two companion rows during the drag: the KE row doubles with T (6.21 → 12.43×10⁻²¹ J — K̄ = (3/2)k_B·T never misses), and v_rms rises by only √2 (517 → 731 m/s); but pressure counts collision impulse AND collision rate, both ∝ v, so P ∝ v² ∝ T — the panel lands on 1995.4 kPa exactly. Option B is the misconception this gate targets: temperature is not a decorative side-channel, it IS the average kinetic energy per molecule, and molecules with doubled energy hammer the sealed walls twice as hard. Option C overcorrects — the square lives inside the speed law (v_rms ∝ √T), not the pressure law. One quiet trap disarmed along the way: the proportionality is to the KELVIN scale — 'the temperature must be expressed in kelvin' (§2.1). Doubling 27 °C to 54 °C doubles nothing: that is only 300 K → 327 K, a 9% rise, and the panel would read 1087.5 kPa, not 1995.4.

Quiz (0/3)

In an isothermal compression from 4 L to 2 L, what happens to pressure?

An isobaric process heats the gas from 300 K to 600 K. How does volume change?

Which of the four processes does the most work on a PV diagram for the same initial and final states?

You can now

  • Apply PV = nRT with the panel's own numbers: at the default state (n=2.0 mol, V=5.0 L, T=300 K) the P row reads 997.7 kPa; seal the box and heat to 600 K and it doubles exactly to 1995.4 kPa — 'at constant volume and number of molecules, the pressure is proportional to the temperature' (OpenStax §2.1, Gay-Lussac's law); squeeze the volume 10→1 L at 300 K and P climbs 498.8→4988.4 kPa, inversely (Boyle); add gas 0.5→5.0 mol and P tracks it linearly, 249.4→2494.2 kPa
  • Connect temperature to microscopic motion: the Avg KE/molecule row is K̄ = (3/2)k_B·T — 'nothing in this equation depends on the molecular mass (or any other property) of the gas… helium and xenon… at the same temperature… have the same average kinetic energy' (§2.2 verbatim) — so at 600 K the row reads 12.43×10⁻²¹ J for ANY gas, while v_rms = √(3RT/M) rises only by √2 (517→731 m/s for nitrogen) because the square lives inside the speed, not the energy
  • Read the Maxwell-Boltzmann board as a probability distribution: its area is pinned to 1 ('∫₀^∞ f(v)dv = 1', §2.4), so as T climbs 300→1000 K the distribution 'is shifted to higher speeds and broadened at higher temperatures' (Fig. 2.16 caption) — wider AND shorter, never taller in place — while the cyan v_rms marker marches 517→944 m/s across the board

Step-by-step

  1. Drag Temperature and watch the pressure row track it linearly in the sealed box — then squeeze Volume and see Boyle's inverse law take over.
  2. Load the presets to audit real states (including the STP button whose own readout disagrees with its label — trust the readout).
  3. The MB distribution board shows the speeds shifting while its area never changes.
  4. Pro mode adds the moles slider.

Key formulas

  • PV=nRT(R=8.314Jmol1K1)PV = nRT \quad (R = 8.314\,\text{J}\cdot\text{mol}^{-1}\text{K}^{-1})Ideal Gas Law
  • W=PΔVW = P\Delta VWork done in isobaric process
  • W=nRTln ⁣(V2V1)W = nRT\ln\!\left(\frac{V_2}{V_1}\right)Work done in isothermal process
  • ΔU=QW\Delta U = Q - WFirst Law of Thermodynamics

Frequently asked questions

In an isothermal compression from 4 L to 2 L, what happens to pressure?
You can work it out this way: use PV = constant for an isothermal process.
An isobaric process heats the gas from 300 K to 600 K. How does volume change?
At constant pressure, V is proportional to T.
Which of the four processes does the most work on a PV diagram for the same initial and final states?
You can work it out this way: compare the area under each curve on the PV diagram.