Ideal Gas Law Calculator

Solve PV = nRT for pressure, volume, moles or temperature with AI-powered step-by-step solutions
n for P = 2.00 atm, V = 5.00 L, T = 300.0 K
Pressure of 0.250 mol in 3.00 L at 27.0 C
Combined gas law: 1.50 L at 1.00 atm and 273.15 K compressed to 0.750 L at 2.50 atm
Volume of 1.00 mol of ideal gas at STP

The Ideal Gas Law

The ideal gas law ties together the four state variables of a gas sample in one equation:

PV=nRTPV = nRT

  • PP — pressure, in atm (or Pa, or kPa).
  • VV — volume, in litres (or m³).
  • nn — amount of gas, in moles.
  • TTabsolute temperature in kelvin, never °C.
  • RR — the universal gas constant. Its numerical value depends entirely on the units you chose: R=0.082057 Latmmol1K1R = 0.082057\ \mathrm{L\,atm\,mol^{-1}K^{-1}} with atm and litres, or R=8.3145 Jmol1K1R = 8.3145\ \mathrm{J\,mol^{-1}K^{-1}} with pascals and cubic metres.

What "ideal" assumes. The molecules are treated as point particles with no volume of their own and no attraction between them, colliding elastically. That is a good description at low pressure and high temperature — roughly, ordinary room conditions for N2\mathrm{N_2}, O2\mathrm{O_2} or He, with errors under about 1%. Near condensation, or above a few tens of atmospheres, real gases deviate and a correction such as the van der Waals equation is needed.

Standard conditions. At STP (0 °C = 273.15 K and 1 bar) one mole of an ideal gas occupies 22.71 L; under the older 1 atm definition it occupies 22.41 L.

Solving for Each Variable

Rearrangements

P=nRTVV=nRTPn=PVRTT=PVnRP = \frac{nRT}{V} \qquad V = \frac{nRT}{P} \qquad n = \frac{PV}{RT} \qquad T = \frac{PV}{nR}

Procedure

  1. Write down the three known quantities with units.
  2. Convert: temperature to kelvin, pressure and volume to units matching your chosen RR (mL → L, kPa → atm by dividing by 101.325, mmHg → atm by dividing by 760).
  3. Substitute and evaluate, carrying units through so they cancel to the unit you expect.
  4. Check plausibility: a mole of gas near room conditions occupies roughly 24 L, so an answer of 0.02 L or 2000 L for one mole signals an arithmetic slip.

The combined gas law

When the amount of gas is fixed and only the conditions change, nn and RR cancel and you never need RR at all:

P1V1T1=P2V2T2\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}

Boyle's Law (TT constant), Charles's Law (PP constant) and Gay-Lussac's Law (VV constant) are each this equation with one variable held fixed.

Significant figures

The result carries the fewest significant figures of the inputs. Because RR is quoted to 5 figures it never limits the answer; a 3-figure pressure does.

Common Mistakes to Avoid

  • Leaving temperature in Celsius. 27 °C is 300.15 K, and using 27 makes the answer wrong by a factor of about 11.
  • Mismatching RR with the units. Using R=0.082057R = 0.082057 with pressure in kPa, or volume in mL, is the single most common source of wrong answers. Pick the RR that matches your units, or convert the units to match RR.
  • Confusing gauge and absolute pressure. PP in PV=nRTPV = nRT is absolute. A tyre gauge reading 2.0 atm means 3.0 atm absolute.
  • Using the combined gas law when moles change. P1V1/T1=P2V2/T2P_1V_1/T_1 = P_2V_2/T_2 is valid only for a sealed, fixed amount of gas; if gas is added, removed or produced by a reaction, go back to PV=nRTPV = nRT for each state.
  • Assuming 22.4 L per mole always. That molar volume belongs to one specific temperature and pressure, not to room conditions.

示例题目

Step 1: Rearrange: n=PVRTn = \dfrac{PV}{RT}
Step 2: Units already match R=0.082057 Latmmol1K1R = 0.082057\ \mathrm{L\,atm\,mol^{-1}K^{-1}} and the temperature is in kelvin
Step 3: n=(2.00)(5.00)(0.082057)(300.0)=10.024.617n = \dfrac{(2.00)(5.00)}{(0.082057)(300.0)} = \dfrac{10.0}{24.617}
Step 4: n=0.40621n = 0.40621 mol, reported to 3 significant figures
Answer: n=0.406n = 0.406 mol

Step 1: Convert the temperature: T=27.0+273.15=300.15T = 27.0 + 273.15 = 300.15 K
Step 2: P=nRTV=(0.250)(0.082057)(300.15)3.00P = \dfrac{nRT}{V} = \dfrac{(0.250)(0.082057)(300.15)}{3.00}
Step 3: Numerator: (0.250)(0.082057)=0.020514(0.250)(0.082057) = 0.020514, and 0.020514×300.15=6.1575 Latm0.020514 \times 300.15 = 6.1575\ \mathrm{L\,atm}
Step 4: P=6.15753.00=2.0525P = \dfrac{6.1575}{3.00} = 2.0525 atm, to 3 significant figures
Answer: P=2.05P = 2.05 atm

Step 1: The amount of gas is fixed, so use P1V1T1=P2V2T2\dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2} — no value of RR is needed
Step 2: Rearrange: T2=T1×P2V2P1V1=273.15×(2.50)(0.750)(1.00)(1.50)T_2 = T_1 \times \dfrac{P_2V_2}{P_1V_1} = 273.15 \times \dfrac{(2.50)(0.750)}{(1.00)(1.50)}
Step 3: 1.8751.50=1.25\dfrac{1.875}{1.50} = 1.25
Step 4: T2=273.15×1.25=341.44T_2 = 273.15 \times 1.25 = 341.44 K, which is 3 significant figures: 341 K
Step 5: In Celsius, 341.44273.15=68.3341.44 - 273.15 = 68.3 °C
Answer: T2=341T_2 = 341 K (about 68 °C)

常见问题

R depends on the units. With pressure in atm and volume in litres, R = 0.082057 L·atm/(mol·K). In SI units (Pa and m³) R = 8.3145 J/(mol·K). With kPa and litres it is 8.3145 L·kPa/(mol·K). The number changes; the physics does not.

At high pressure and low temperature. The model ignores molecular volume and intermolecular attraction, so it fails as molecules are squeezed close together or slowed near condensation. Below a few atmospheres and well above the boiling point, errors are usually under 1%.

PV = nRT describes one state of a gas and can find the number of moles. The combined gas law, P1V1/T1 = P2V2/T2, compares two states of the same sealed sample; n and R cancel, so it cannot give you moles but it also does not need R.

Find n = PV/(RT), then divide the measured mass by n: M = m/n = mRT/(PV). Equivalently, with density d = m/V, M = dRT/P. Both require the temperature in kelvin and a value of R matching your pressure and volume units.

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