Ideal Gas Law Calculator

Solve the ideal gas equation \(PV=nRT\) for pressure, volume, or temperature using your known gas conditions.

Configuration

Pick what to solve for, then enter the remaining variables.

Known Values

Result

Practical Guide

Ideal Gas Law Calculator

This guide shows how to use the Ideal Gas Law Calculator correctly, interpret the outputs, and spot the common pitfalls. You’ll see when \(PV=nRT\) is valid, how to pick the right inputs and units, and how to sanity-check results with worked examples. The goal is simple: get reliable pressure, volume, or temperature estimates without forcing you to re-derive thermodynamics every time.

6–8 min read Updated 2025 Thermodynamics / Fluids

Quick Start

The calculator is based on the Ideal Gas Law: \[ PV = nRT \] where \(P\) is absolute pressure, \(V\) is volume, \(n\) is amount of gas (moles), \(R\) is the universal gas constant, and \(T\) is absolute temperature. Here’s the fastest safe workflow.

  1. 1 Choose what you want to compute in Solve For: Pressure \(P\), Volume \(V\), or Temperature \(T\). The calculator will hide that row automatically.
  2. 2 Enter the known variables. Use realistic engineering magnitudes (avoid “placeholder” zeros).
  3. 3 Confirm that your pressure input is absolute, not gauge. If you measured gauge pressure, convert it first: \[ P_{\text{abs}} = P_{\text{gauge}} + P_{\text{atm}} \] with \(P_{\text{atm}} \approx 101.325\ \text{kPa} = 14.7\ \text{psi}\).
  4. 4 Make sure temperature is in an absolute scale. If you input °C or °F, the calculator converts internally to K or °R: \[ T(K)=T(^{\circ}C)+273.15,\quad T(^{\circ}R)=T(^{\circ}F)+459.67 \]
  5. 5 Pick result units under Result Units (kPa, psi, m³, L, °C, K, etc.) to match your application.
  6. 6 Review Quick Stats and the equation banner. If \(PV\) and \(nRT\) don’t agree within rounding, you likely have a unit or absolute/gauge mistake.
  7. 7 Use Show Steps to verify the rearrangement and substituted values before you stamp a calculation into a report.

Tip: If your gas is near atmospheric pressure and room temperature, the ideal gas law is usually accurate to within a few percent for many common gases.

Watch out: At very high pressures, very low temperatures, or near condensation, real-gas effects can make \(PV=nRT\) underestimate or overestimate the true state.

Choosing Your Method

Engineers use the ideal gas law in a few recurring ways. The calculator supports the direct \(PV=nRT\) form, but you should pick the framing that best matches what you actually measured.

Method A — Direct \(PV=nRT\) with moles

Use this when the amount of gas is naturally expressed in moles (lab work, chemical processes, reaction stoichiometry).

  • Matches the universal constant \(R=8.314462618\ \text{J/(mol·K)}\).
  • Cleanest algebra; easy to rearrange for any variable.
  • Ideal for mixtures when total moles are known.
  • Field measurements are often mass-based, not mole-based.
  • Requires a molar mass conversion if you start from kilograms or pounds.
\(\;PV=nRT\;\)

Method B — Specific gas constant form \(PV=mR_sT\)

Best when you know the gas mass \(m\) instead of moles. Convert first, then still use the calculator.

  • Natural for HVAC, compressed air systems, and engine cylinders.
  • Mass is easy to measure or infer from flow meters.
  • Requires gas identity and molar mass \(M\).
  • You must convert to moles using \(n=m/M\).
\(\;n=\dfrac{m}{M}\;\Rightarrow\;PV=mR_sT\)

Method C — Real-gas correction with compressibility \(Z\)

Use this when pressures are high or temperatures approach saturation. The calculator is still useful as a baseline.

  • Captures real behavior in tanks, pipelines, and cryogenic work.
  • Reduces systematic error versus ideal assumptions.
  • Requires charts or EOS software to find \(Z\).
  • More work than ideal gas; often overkill below ~5–10 bar.
\(\;PV=ZnRT\;\) with \(Z\approx1\) for ideal gases

In practice, apply Method A/B for everyday engineering sizing and sanity checks, and step up to Method C only when the ideal assumption is a known risk to safety or performance.

What Moves the Number the Most

Because the ideal gas law is multiplicative, a small mistake in any dominant variable scales right into your result. These are the main “levers” that swing the output:

Absolute temperature \(T\)

Temperature must be absolute. Forgetting the \(+273.15\) or \(+459.67\) shift is the #1 source of huge errors. Pressure and volume are directly proportional to \(T\).

Pressure reference (absolute vs gauge)

\(P\) in \(PV=nRT\) is absolute. A 50 psi gauge reading is ~64.7 psia. Missing this offset makes your computed volumes or temperatures too low.

Amount of gas \(n\)

If \(n\) comes from mass, convert with molar mass \(M\): \[ n=\frac{m}{M} \] A wrong molar mass (air vs nitrogen vs CO₂) gives a proportional error.

Volume \(V\)

For tanks and vessels, volume uncertainty can dominate. Verify internal free volume (subtract liners, piping dead-legs, or liquid fill).

Unit consistency

The calculator converts for you, but mixed assumptions still happen: e.g., using liters with psi without selecting correct units, or entering ft³ but thinking in gallons.

Real-gas deviation \(Z\)

Near critical points, high pressures, or very low temperatures, \(Z\neq1\). The ideal calculation becomes a first guess that you should correct.

Worked Examples

These examples mirror typical use cases. Follow the same steps in the calculator to validate your own numbers.

Example 1 — Find Pressure in a Sealed Cylinder

  • Known volume: \(V = 0.050\ \text{m}^3\) (50 L tank)
  • Known amount: \(n = 2.00\ \text{mol}\)
  • Known temperature: \(T = 35^{\circ}\text{C}\)
  • Solve for: Pressure \(P\)
1
Convert to absolute temperature: \[ T = 35 + 273.15 = 308.15\ \text{K} \]
2
Rearrange the ideal gas law: \[ P = \frac{nRT}{V} \]
3
Substitute values: \[ P = \frac{(2.00)(8.314)(308.15)}{0.050} = 102{,}500\ \text{Pa} \]
4
Convert to kPa or atm: \[ P \approx 102.5\ \text{kPa} \approx 1.01\ \text{atm} \]

Interpretation: This is basically atmospheric pressure, which makes sense: 2 mol in a 50 L vessel at warm room temperature isn’t highly compressed.

Example 2 — Find Required Volume for Compressed Air

  • Known pressure (absolute): \(P = 600\ \text{kPa}\) (≈ 5.9 atm)
  • Known amount: \(n = 10.0\ \text{mol}\)
  • Known temperature: \(T = 20^{\circ}\text{C}\)
  • Solve for: Volume \(V\)
1
Absolute temperature: \[ T = 20 + 273.15 = 293.15\ \text{K} \]
2
Rearranged form: \[ V = \frac{nRT}{P} \]
3
Substitute (convert pressure to Pa): \[ P = 600\ \text{kPa} = 600{,}000\ \text{Pa} \] \[ V = \frac{(10.0)(8.314)(293.15)}{600{,}000} = 0.0406\ \text{m}^3 \]
4
Convert to liters: \[ V = 0.0406\ \text{m}^3 \times 1000 = 40.6\ \text{L} \]

Interpretation: A ~41 L vessel at 600 kPa abs holds 10 mol of ideal gas at 20°C. If this were an air receiver, you would check whether real-gas effects are small at ~6 atm (they usually are for air).

Common Layouts & Variations

The same equation shows up across many disciplines. The table below summarizes common configurations and the extra checks you should do before trusting a pure ideal-gas result.

Configuration / Use CaseTypical Known InputsWhy Ideal Gas Works (or Doesn’t)Practical Notes
Laboratory flask or syringe\(n, V, T\)Low pressure, far from saturation → \(Z\approx1\)Use absolute temperature; small volumes make unit errors obvious.
Compressed air receiver\(P, V, T\) → solve for \(n\) or \(m\)Air is near-ideal below ~10 bar at ambient \(T\)Confirm pressure is absolute; verify free internal volume.
Engine cylinder (intake/combustion)\(P, V, T\) varyingReasonable first-pass for intake statesDuring combustion, gases may deviate; use ideal as baseline only.
High-pressure gas bottle\(P, n, T\) → solve for \(V\)At high \(P\), real-gas effects matterApply \(Z\) correction if \(P\gtrsim 20\) bar or near critical \(T\).
Cryogenic or near-condensing gasAny three variablesOften non-ideal; phase change riskDo not rely on ideal gas law alone—check saturation curves/EOS.
  • Check that temperature is truly uniform (no hot spots or stratification).
  • Use the gas identity you actually have, not “air” by default.
  • Verify whether your pressure gauge reads absolute or gauge.
  • For mixtures, total moles matter: \(n=\sum n_i\).
  • If results look extreme, re-enter values in SI to validate units.
  • Consider \(Z\) for high-pressure sizing or safety analysis.

Specs, Logistics & Sanity Checks

The ideal gas law is a model. It’s extremely useful, but only if you feed it defensible inputs. Before finalizing a design or troubleshooting a system, do these engineering checks.

Pressure sanity checks

  • Convert gauge to absolute: add local atmospheric pressure.
  • At altitude, \(P_{\text{atm}}\) is lower—don’t assume sea level.
  • If computed \(P\) exceeds equipment rating, stop and reassess assumptions.

Volume sanity checks

  • Use internal free volume (subtract liquid, liners, baffles, or dead-legs).
  • For irregular vessels, validate volume geometrically or by water-fill tests.
  • Be consistent: ft³, in³, L, and m³ differ by orders of magnitude.

Temperature sanity checks

  • Measure where the gas is, not where the pipe wall is.
  • Correct for transient heating/cooling if compression or expansion is rapid.
  • Never input °C/°F thinking it is already absolute.

Rule of thumb: If your operating point is below ~5 bar and above ~0°C for common gases (air, N₂, O₂), the ideal model is typically fine for sizing and troubleshooting. If you’re near the limits of a pressure vessel, near saturation, or in cryogenic ranges, use the calculator as a first pass and then confirm with a real-gas method.

Frequently Asked Questions

Do I need to use absolute pressure in the Ideal Gas Law Calculator?
Yes. The ideal gas law uses absolute pressure. If you have gauge pressure, add atmospheric pressure first: \(P_{\text{abs}} = P_{\text{gauge}} + P_{\text{atm}}\). Using gauge pressure directly will under-predict results.
Why does the calculator require absolute temperature?
Because \(T\) in \(PV=nRT\) is absolute. Celsius and Fahrenheit are offset scales. The calculator converts °C to K and °F to °R internally, but you must enter a real physical temperature (not a temperature difference).
Can I use mass instead of moles?
You can, but convert mass to moles first using the molar mass \(M\): \(n=m/M\). For example, 0.029 kg/mol for dry air. Then enter \(n\) in mol, kmol, or lbmol.
At what conditions does the ideal gas law stop being accurate?
Accuracy drops at high pressures, very low temperatures, or near condensation/critical points. As a rough guide, many gases are close to ideal below about 5–10 bar at ambient temperatures, but you should use a compressibility factor \(Z\) if deviations matter.
What gas constant does this calculator use?
It uses the universal gas constant \(R=8.314462618\ \text{J/(mol·K)}\). If you’re using a specific gas constant \(R_s\), convert to moles or use \(n=m/M\) so the universal \(R\) applies.
Why are my results off by about 14.7 psi (or 101 kPa)?
That’s almost always a gauge vs absolute issue. A pressure gauge reads relative to atmosphere. Adding \(P_{\text{atm}}\) corrects the offset and aligns the ideal gas law with your measurement.
How can I sanity-check my calculation quickly?
Check proportionality: if you double \(T\) (in K), \(P\) or \(V\) should roughly double, all else equal. Also compare the Quick Stats: \(PV\) and \(nRT\) should match within rounding.
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