Ideal Gas Law Calculator: Solve P, V, n, T (PV=nRT)

If you're plugging numbers into an ideal gas law calculator and the pressure comes out as 0.003 atm for a balloon at room temperature, something went wrong with your units. PV = nRT has four variables—pressure, volume, moles, and temperature—and you can solve for whichever one you're missing. The algebra is just division: P = nRT/V, V = nRT/P, n = PV/(RT), T = PV/(nR). The hard part isn't the math. It's making sure every value is in the right unit before you touch the calculator.

The most common error is using Celsius for temperature. PV = nRT requires absolute temperature in kelvin. Plug in 25 (meaning 25 °C) instead of 298 K, and your answer is off by a factor of about 12. The equation breaks entirely at 0 °C if you forget to convert, because T = 0 makes the right side zero regardless of P, V, or n. Kelvin exists precisely to avoid this—its zero point is absolute zero, where molecular motion stops.

Once you solve for the unknown, check the result against intuition. One mole of gas at STP (273.15 K, 1 atm) occupies 22.4 L. If your answer gives 0.5 mol occupying 200 L at 1 atm and 300 K, that's suspiciously large—roughly 4× what you'd expect. Sanity checks like this catch unit errors, misplaced decimals, and wrong R values before you submit.

The gas constant R has the same physical meaning in every unit system, but its numerical value changes depending on which pressure and volume units you use. R = 0.08206 L·atm/(mol·K) only works when P is in atm and V is in liters. Use kPa for pressure with that R, and your answer is wrong by a factor of 101.325. Use mL instead of L, and you're off by 1000.

Here are the R values you'll encounter most often: 0.08206 L·atm/(mol·K) for atm + L, 8.314 J/(mol·K) or equivalently 8.314 kPa·L/(mol·K) for kPa + L or Pa + m³, and 62.36 L·mmHg/(mol·K) for mmHg + L. Pick the R that matches your given units. If your problem gives pressure in bar, use R = 0.08314 L·bar/(mol·K). Dimensional analysis catches mismatches: write out every unit and cancel. If the result doesn't reduce to the target unit, your R is wrong.

Volume trips people up less often, but watch for mL vs L and cm³ vs m³. One cubic meter is 1000 liters, not 100. If your textbook gives volume in cm³ (which equals mL), convert to L before using R = 0.08206. And never use Fahrenheit—convert to Celsius first, then add 273.15 to get kelvin. Two conversions, zero shortcuts.

When the amount of gas stays fixed (sealed container, no leaks), you can compare two states without knowing n or R at all. The combined gas law P₁V₁/T₁ = P₂V₂/T₂ falls out of PV = nRT because nR cancels on both sides. You fill in the initial state (subscript 1), the final state (subscript 2), and solve for the one missing value.

This is what you use for "before and after" problems: a gas at 2.0 atm, 300 K, 5.0 L is heated to 450 K at constant pressure—find the new volume. Constant pressure means P₁ = P₂, so they cancel: V₁/T₁ = V₂/T₂. V₂ = V₁ × T₂/T₁ = 5.0 × (450/300) = 7.5 L. If pressure also changes, keep all six variables and solve algebraically.

The combined law also contains Boyle's law (constant T: P₁V₁ = P₂V₂), Charles's law (constant P: V₁/T₁ = V₂/T₂), and Gay-Lussac's law (constant V: P₁/T₁ = P₂/T₂) as special cases. You don't need to memorize each separately—just use P₁V₁/T₁ = P₂V₂/T₂ and cancel whatever's held constant. Label your states clearly ("State 1: before heating", "State 2: after heating") so you never mix initial and final values.

Replace n with m/M (mass divided by molar mass) in PV = nRT, and rearrange: d = PM/(RT), where d is gas density in g/L. This is useful in two directions. Given a known gas and conditions, calculate its density. Or given measured density at known T and P, solve for molar mass: M = dRT/P. The second version is how chemists historically identified unknown gases.

At STP (273.15 K, 1 atm), any ideal gas has density d = M/22.414 g/L. For O₂ (M = 32.00 g/mol), d = 32.00/22.414 = 1.428 g/L. For He (M = 4.003 g/mol), d = 0.1786 g/L. Helium's density is about 7% that of air (average M ≈ 29 g/mol), which is why helium balloons float. If your density calculation gives a heavier-than-air number for helium, check your M or your units.

When using d = PM/(RT) with R = 0.08206, pressure must be in atm, M in g/mol, and T in kelvin. The density comes out in g/L. If you want kg/m³ instead, multiply by 1 (since 1 g/L = 1 kg/m³). But if you accidentally use R = 8.314 (Pa·m³ units), then P must be in Pa, M in kg/mol, and density comes out in kg/m³. Mixing unit systems gives nonsense—always trace through dimensional analysis.

Which R should I use? Match R to whatever pressure and volume units you have. R = 0.08206 L·atm/(mol·K) is the standard for general chemistry. If your pressure is in kPa, use R = 8.314 kPa·L/(mol·K). If mmHg (Torr), R = 62.36 L·mmHg/(mol·K). If bar, R = 0.08314 L·bar/(mol·K). Temperature is always kelvin regardless of which R you pick.

Does R = 8.314 J/(mol·K) work for PV = nRT? Yes, but only when P is in pascals and V is in m³, because 1 J = 1 Pa·m³. This version appears in thermodynamics and physical chemistry more often than in general chemistry. If you use it with liters and atmospheres, you'll be off by 101,325.

Can I convert units instead of switching R? Absolutely. If your problem gives P = 152 kPa, you can convert to atm (152/101.325 = 1.50 atm) and use R = 0.08206. Or leave P in kPa and use R = 8.314. Either path gives the same answer. The only rule: every unit in P, V, T must match R. One mismatch and the number is garbage.

What about the calorie version? R = 1.987 cal/(mol·K) shows up in thermochemistry when energies are given in calories. It's the same R, just in calorie-based energy units. You wouldn't use this for PV = nRT gas calculations—it's for ΔG = −RT ln K when ΔG is in cal/mol. Using the wrong R in thermodynamic equations is just as destructive as using the wrong R in gas law equations.

When does the ideal gas law fail? At high pressures (above ~10 atm) and low temperatures (near the boiling point of the gas). Under these conditions, gas molecules are close enough that their physical volume and intermolecular attractions matter. The van der Waals equation [P + a(n/V)²](V − nb) = nRT corrects for both effects using gas-specific constants a (attraction) and b (molecular volume). For typical homework problems at 1–5 atm and 273–400 K, the ideal gas law is accurate to within a few percent.

Why can't I use PV = nRT for liquids? The equation assumes freely moving particles with negligible volume and no persistent interactions. Liquids have strong intermolecular forces and essentially fixed density. Water at 25 °C and 1 atm has a molar volume of 0.018 L/mol, but PV = nRT predicts 24.5 L/mol for the same conditions. That's off by a factor of 1400. The ideal gas law applies only to the gas phase.

Does PV = nRT work for gas mixtures? Yes. Each gas in a mixture independently obeys PV = nRT using its own moles: P_iV = n_iRT. The total pressure is the sum of all partial pressures (Dalton's law). The total moles give total pressure: P_totalV = n_totalRT. This works because ideal gas particles don't interact, so each species behaves as if the others aren't there.

Problem: A rigid 12.0 L steel cylinder contains 0.50 mol of N₂ at 22 °C. Find the pressure inside the cylinder.

The sanity check confirms the result. If we had forgotten to convert 22 °C to kelvin and used T = 22 directly, we'd get P = (0.50)(0.08206)(22)/12.0 = 0.075 atm—a near vacuum inside a sealed cylinder at room temperature, which is physically absurd. That's why the kelvin conversion matters every single time.