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🧴 Calculators

Ideal Gas Volume PV nRT

Computes volume V of an ideal gas from n, T (K) and P (atm) via PV = nRT.

Ideal gas volume: V = nRT / P

Rearrange the ideal gas law PV = nRT and you get V = nRT/P, the volume that n moles of an ideal gas take up at temperature T (kelvin) and pressure P. With R = 0.082057 L·atm/(mol·K), the classical "CNTP/STP" reference of 273.15 K and 1 atm gives 22.4 L for 1 mol of any ideal gas. Switch to the post-1982 IUPAC STP (273.15 K, 100 kPa) and the molar volume comes out to 22.7 L instead. The law tracks real gases well at low pressure and high temperature, but it starts to drift once intermolecular forces become significant. When that happens, reach for the van der Waals or Redlich–Kwong equations.

Applications

Teaching general chemistry, sizing industrial O₂, N₂ and H₂ storage tanks, working out the stoichiometry of gas-phase reactions like combustion or ammonia synthesis, HVAC and process engineering, atmospheric science, and quick lab estimates of how much gas was evolved or consumed.

FAQ

CNTP, STP or NTP — what is the difference? The old "CNTP" used in Brazil and the pre-1982 STP both ran on 273.15 K and 1 atm, giving 22.4 L/mol. IUPAC's modern STP shifted to 273.15 K and 100 kPa (22.7 L/mol). NTP, meanwhile, usually means 293.15 K and 1 atm (24.0 L/mol).

Does the gas identity matter? For an ideal gas, no. One mole of He, O₂ or CO₂ takes up the same volume at the same P and T. Identity only comes into play once you start applying non-ideal corrections.

Why must temperature be in kelvin? The law comes out of kinetic theory, which is built on absolute temperature. Plug in °C and you'd hit zero or negative values at low T, and the equation falls apart.

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Ideal Gas Law: Solve for Pressure (PV = nRT)

Enter moles, temperature in kelvin and volume in liters to get P = nRT/V in atm, with R = 0.08206 L atm per mol per kelvin.

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Cutting Force by the Kienzle Equation

Computes the main cutting force with the Kienzle equation, F_c = k_c1.1 · b · h^(1 − m_c), where k_c1.1 is the tabulated specific cutting force of the workpiece material for a reference chip section of 1 mm × 1 mm, b is the chip width and h the chip thickness, and m_c is the exponent describing the size effect. That is exactly where it differs from the direct calculation F_c = k_s·b·h: the latter treats specific pressure as a material constant, while Kienzle embeds the experimental fact that thin chips cost far more force per unit area, because the cutting edge radius stops being negligible next to the chip thickness. With k_c1.1 = 1500 N/mm² and m_c = 0.26, a 0.2 mm thick chip works at 2279 N/mm², 52 % above the tabulated value — which is why very low feeds raise the power spent per cubic millimetre removed, and the tool wear with it, instead of saving them — even though the absolute force falls. Since k_c1.1 carries a hidden millimetre raised to m_c, the equation is not dimensionally pure: thickness and width have to be entered in millimetres, and switching units is off by orders of magnitude. Enter the specific force k_c1.1, the exponent m_c, the chip width and the chip thickness.

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PLA Filament by Volume and Density

Estimates PLA filament mass and length needed from part volume and density.

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Sauce Reduction Ratio Calculator

Starting and target volume in ml give the reduction factor and the share of liquid boiled off: 1000 ml down to 250 ml is 4.00x, or 75 percent.

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Theoretical Methane Yield (Buswell)

Computes the theoretical methane yield of a substrate with the Buswell equation, which closes the stoichiometric balance of anaerobic digestion of a CₙHₐO_bN_c compound into methane, carbon dioxide and ammonia: each mole of substrate yields (4n + a − 2b − 3c) ÷ 8 moles of methane, and dividing that by the molar mass and multiplying by the 22.414 L/mol molar volume gives the yield in litres of methane per gram at normal conditions, 0 °C and 1 atm. The less oxygen the molecule already carries, the more reduced it is and the more methane it yields: cellulose and glucose land at 0.41 and 0.37 L/g with 50% methane in the biogas, while a fat such as tristearin exceeds 1.02 L/g and reaches 71% methane — the methane fraction of the biogas is exactly (4n + a − 2b − 3c) ÷ 8n, since all the substrate carbon leaves either as methane or as carbon dioxide. The value is a thermodynamic ceiling, not a design forecast: in practice a digester delivers 60% to 80% of it, because part of the substrate becomes bacterial biomass and part never becomes accessible to the enzymes within the available retention time. Enter the number of carbon, hydrogen, oxygen and nitrogen atoms in the substrate's empirical formula.

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Tunnel Volume Loss

Calculate the volume loss of a tunnel excavation, VL = Vs ÷ (π·D²/4)·100, the percentage ratio between the settlement trough volume per metre Vs (m³/m) and the excavated cross-section area (from diameter D). Volume loss quantifies how much soil 'disappeared' relative to the theoretical tunnel volume — caused by face relaxation, overexcavation, tail-gap closure behind the TBM shield and consolidation. It is the key control parameter for urban excavation: well-run EPB/slurry TBMs achieve 0.5-1.5% in soils; values above 2-3% indicate problems and excessive settlement. Enter the trough volume and the tunnel diameter.

The results provided by this tool are for general informational and educational purposes only and do not constitute professional, financial, medical, legal, tax or accounting advice. Always confirm important decisions with a qualified professional and official sources.