SOLVETUTORMATH SOLVER

Instrument MI-03-090 · Physics

Combined Gas Law Calculator

Squeeze a trapped gas and heat it at the same time and pressure moves from both effects together — one ratio folds Boyle's, Charles's and Gay-Lussac's laws into a single identity.

Instrument MI-03-090
Sheet 1 OF 1
Rev A
Verified
Type 03 — Thermodynamics SER. 2026-03090

Final pressure

236.425000 kPa

P₂ = P₁V₁T₂ ⁄ (V₂T₁)

The working Every figure verified twice
  1. P2 = 101325·0.01·350 ⁄ (0.005·300) = 236,425.000000
Worksheet log
  1. No entries yet — change an input to log a scenario.

How this instrument works

The combined gas law is what's left of the ideal gas law PV = nRT once you cancel the two things that don't change between two snapshots of the same trapped gas: the amount of substance n and the gas constant R. What survives is P₁V₁ ⁄ T₁ = P₂V₂ ⁄ T₂ — pressure times volume divided by absolute temperature holds the same value before and after, provided no gas is added, removed, or converted into something else along the way. That ratio is a state function: it depends only on where the gas started and where it ended up, never on the road taken between those two points, whether that road was a slow squeeze, a sudden heat spike, or both running together.

One way to see why the formula multiplies rather than adds its pieces is to walk it in two steps. First compress the gas at constant temperature T1, which is Boyle's law: an intermediate pressure Pi = P1V1 ⁄ V2. Then, holding that new volume V2 fixed, heat the gas from T1 up to T2, which is Gay-Lussac's law: Pi rises to P2 = Pi · T2 ⁄ T1. Substitute the first result into the second and the intermediate pressure cancels out of the middle, leaving P2 = P1V1T2 ⁄ (V2T1) — exactly the formula this instrument solves. The gas never actually has to be compressed first and heated second; the arithmetic holds whichever order the two changes really happened in, or if they happened at once.

Two assumptions carry all of this and both can fail quietly. The amount of gas has to stay fixed — a reacting mixture, a leaking seal, or a headspace absorbing vapour all break the identity even though the arithmetic keeps returning a number. And the gas has to behave close to ideally: dilute, well above its condensation point, molecules that mostly ignore one another. Compress a real gas hard enough, or chill it toward where it would liquefy, and measured pressure drifts away from what this ratio predicts — by a few percent at ordinary engineering pressures, and by far more near a phase change.

P1V1T1=P2V2T2\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}P2=P1V1T2V2T1P_2 = \frac{P_1 V_1 T_2}{V_2 T_1}V2=P1V1T2P2T1V_2 = \frac{P_1 V_1 T_2}{P_2 T_1}
P₁, P₂ — absolute pressure before and after, in pascals (Pa) or the chosen unit · V₁, V₂ — volume before and after, in cubic metres (m³) or the chosen unit · T₁, T₂ — absolute temperature before and after, kelvin (K) only. Amount of gas stays fixed throughout; pressures must be absolute, never gauge.
  • Enter Initial pressure and Initial volume for the gas's starting state. Each carries its own unit menu — Pa, kPa or atm; ml, l or m³ — and the instrument normalizes everything to SI before solving.
  • Enter Initial temperature, K directly in kelvin; this field has no unit menu, so convert first if your reading is in Celsius (K = °C + 273.15) or Fahrenheit.
  • Enter Final volume and Final temperature, K for the state you want the pressure at — the volume the gas is squeezed or let out to, and the kelvin temperature it reaches there.
  • Read Final pressure, then flip its unit menu to whichever pressure unit you need to check against a gauge, spec sheet, or relief-valve rating.

Worked example — sizing a reactor headspace relief valve

A process engineer seals a 10-litre nitrogen headspace above a batch reactor at Initial pressure 101.325 kPa, one atmosphere, and Initial temperature, K of 300 — a mild 27 °C start. As the reaction runs, incoming liquid reagent squeezes that headspace down to a Final volume of 5 litres, while the exothermic reaction itself carries the gas up to a Final temperature, K of 350, about 77 °C. Substituting into P₂ = P₁V₁T₂ ⁄ (V₂T₁): P₂ = 101325 × 0.01 × 350 ⁄ (0.005 × 300) = 236425 Pa, or 236.425 kPa.

Halving the volume alone, in the style of Boyle's law, would have doubled pressure to 202.65 kPa; folding in the temperature climb from 300 K to 350 K — a factor of 7 ⁄ 6 — pushes it further to 236.425 kPa, close to 2.33 atmospheres absolute. That combined figure is what actually sizes the relief valve and the vessel's pressure rating; reading off compression or heating alone would have under-specified the hardware. It is also why the calculator insists on kelvin rather than accepting 27 °C and 77 °C directly — run those Celsius figures through the same ratio unconverted and the multiplier comes out badly wrong instead of the true 7 ⁄ 6.

Questions

Why does this law need three variables changing at once?

Because real processes rarely hold anything constant. Boyle's law needs steady temperature, Charles's law needs steady pressure, Gay-Lussac's law needs steady volume — useful special cases, but a compressor stroke, a reactor headspace, or a weather balloon's climb typically moves two or three quantities together. The combined form removes that restriction: feed it any two complete states of the same trapped gas and it connects them directly, without caring which quantity moved first or by how much.

Does it matter what order the compression and heating happen in?

No. Pressure, volume and temperature are state functions — their combined ratio depends only on the starting and ending values, not on the path between them. Compress first and heat second, heat first and compress second, or change both at once: any route between the same two states returns the same final pressure. That path-independence is exactly what separates a state function from something like heat or work, which do depend on the process.

What happens if I plug in a Celsius or Fahrenheit reading by mistake?

The result comes out wrong, often badly, because the ratio only means what it claims when zero actually means zero. Celsius and Fahrenheit put their zeros somewhere arbitrary, so 20 °C is not physically half of 40 °C the way 293.15 K is a genuine fraction of 313.15 K. Convert first: kelvin = °C + 273.15, or kelvin = (°F + 459.67) ⁄ 1.8.

Why must the amount of gas stay the same between the two states?

Because the derivation cancels n and R straight out of PV = nRT, which only holds if neither one changed. Adding gas, venting it through a relief valve, or letting it react or condense between readings changes n mid-calculation, and the instrument has no way to know that happened — it will still hand back a number, just not a physically meaningful one. Use it only across a genuinely closed, fixed-mass sample of gas.

How is this different from just using the ideal gas law directly?

The ideal gas law, PV = nRT, needs the amount of substance n and the gas constant R stated explicitly. The combined form comes from writing that equation at two different moments and dividing one by the other: n and R cancel completely, so you never need to know how many moles of gas you actually have. That is a real convenience whenever two full states are known but the mass or mole count connecting them is not.

At what point does a real gas stop matching this ratio?

Mainly near condensation and at high pressure, where molecules start occupying real volume and pulling on one another instead of behaving as independent points. Dry air, nitrogen and similar gases well above their boiling point track this formula within a fraction of a percent up to several bar; push past roughly 10 bar, or chill a vapour toward its dew point, and measured pressure drifts below what the ideal ratio predicts.

References