The Ideal Gas Law PV = nRT理想气体方程 PV = nRT

Big Idea · Structure and Properties (SAP)AP Science Practice 5 · Mathematical RoutinesAP link · Unit 3~40 min

Learning goals

  • Use the ideal gas law, PV = nRT, to find pressure, volume, moles, or temperature, choosing the value of R that matches the units
  • Find a gas's molar mass from its mass, pressure, volume, and temperature, or from its density
  • Explain what an ideal gas is, and when real gases deviate from ideal behavior

Before you start

  • The combined gas law and Avogadro's law (2.8.3)

How much does the air in a classroom weigh?

You can’t see or feel the air in a classroom, so it seems weightless. But measure the room’s length, width, and height, read the thermometer and the barometer, and you can calculate how many moles and how many kilograms of air are inside. The tool is one equation that ties P, V, n, and T together.

The ideal gas law

Combine Boyle’s law, Charles’s law, Gay-Lussac’s law, and Avogadro’s law (V ∝ n), and you get

PV=nRTPV = nRT

This is the ideal gas law理想气体状态方程. R is the gas constant摩尔气体常数, the same for every gas. Its value depends on the units:

Units of PUnits of VR
atmL0.08206 L·atm/(mol·K)
torr (mmHg)L62.36 L·torr/(mol·K)
kPaL8.314 L·kPa/(mol·K)
Pam³8.314 J/(mol·K)

These are the values on the AP equation sheet (plus the kPa version). The last two rows share a number because 1 Pa·m³ = 1 J, and 1 kPa·L is also 1 J. The exact value is 8.314 462 618 J/(mol·K). T is always in kelvins.

Substituting n = m/M gives the molar mass or the density of a gas:

M=mRTPVd=PMRTM = \frac{mRT}{PV} \qquad d = \frac{PM}{RT}

What makes a gas “ideal”?

An ideal gas is a model in which the particles themselves take up no volume and don’t attract or repel each other. At ordinary temperatures and pressures, most gases come very close, and PV = nRT is accurate. At very high pressure the particles are crowded together, so their own volume matters; at very low temperature they move slowly enough for attractions to matter. In both cases real gases deviate noticeably from PV = nRT.

Chinese chemistry textbooks don’t use PV = nRT; students meet the ideal gas equation in physics, written as pV/T = constant. Chinese chemistry problems usually give standard conditions and use Vm = 22.4 L/mol. PV = nRT is more general: it works at any temperature and pressure, and 22.4 L/mol is just its value at 0 °C and 1 atm.

This lesson uses the atomic masses on the AP periodic table (H 1.008, C 12.01, N 14.01, O 16.00, S 32.06, Cl 35.45).

Choose an unknown and fill in the other three quantities to see how R is chosen and how the units cancel. The last button finds a molar mass from the mass of a gas sample and suggests which gas it might be:

Try it: ideal gas law (PV = nRT) solver
Solve for

    Example 1: the air in a classroom

    A classroom is 10.0 m long, 8.00 m wide, and 3.00 m high. The air is at 25 °C and 1.00 atm. How many moles of air are in the room? Taking the average molar mass of air as 29.0 g/mol, what is its mass?

    Step-by-step solution

    1. Choose R. With P in atm and V in liters, R = 0.08206 L·atm/(mol·K), so convert the volume to liters and the temperature to kelvins first.

    2. Volume in liters. 1 m³ = 1000 L:

      V=10.0×8.00×3.00 m3=240 m3=2.40×105 L\begin{aligned} V &= 10.0 \times 8.00 \times 3.00\ \text{m}^3 \\ &= 240\ \text{m}^3 \\ &= 2.40 \times 10^{5}\ \text{L} \end{aligned}

    3. Temperature in kelvins. T = 25 + 273 = 298 K. The temperature is given to the ones place, so adding 273 is enough (2.8.2).

    4. Moles. From PV = nRT, n = PV/RT:

      n=PVRT=1.00×2.40×1050.08206×298=9814≈9.81×103 mol\begin{aligned} n &= \frac{PV}{RT} \\ &= \frac{1.00 \times 2.40 \times 10^{5}}{0.08206 \times 298} \\ &= 9814 \approx \boxed{9.81 \times 10^{3}\ \text{mol}} \end{aligned}

      Units: atm·L ÷ (L·atm/(mol·K) × K) = mol.

    5. Mass. Keep the unrounded 9814 mol and round only at the end:

      m=nM=9814×29.0=2.85×105 g\begin{aligned} m &= nM \\ &= 9814 \times 29.0 \\ &= \boxed{2.85 \times 10^{5}\ \text{g}} \end{aligned}

      That’s about 285 kg, heavier than four adults.

    6. A common mistake. Substituting V = 240 (in m³) gives an n that is 1000 times too small. Before substituting, check which units R uses.

    Example 2: identifying a gas

    A 250-mL flask filled with an unknown gas at 25 °C and 1.00 atm contains 0.655 g of the gas. What is its molar mass, and what might the gas be?

    Step-by-step solution

    1. The plan. M = m ÷ n. The mass is given; find the moles from PV = nRT.

    2. Units. V = 250 mL = 0.250 L and T = 298 K.

    3. Moles.

      n=PVRT=1.00×0.2500.08206×298=1.022×10−2 mol\begin{aligned} n &= \frac{PV}{RT} \\ &= \frac{1.00 \times 0.250}{0.08206 \times 298} \\ &= 1.022 \times 10^{-2}\ \text{mol} \end{aligned}

      Keep an extra digit for now.

    4. Molar mass.

      M=mn=0.6551.022×10−2=64.1 g/mol\begin{aligned} M &= \frac{m}{n} \\ &= \frac{0.655}{1.022 \times 10^{-2}} \\ &= \boxed{64.1\ \text{g/mol}} \end{aligned}

    5. Compare. SO₂ has a molar mass of 32.06 + 2(16.00) = 64.06 g/mol, the closest match. The gas is probably SO₂.

    6. Another route. Combine the two steps into M = mRT ÷ (PV). The numbers and the answer are the same.

    Practice: which value of R should you use?
    1. P in atm, V in L
    2. P in kPa, V in L
    3. P in mmHg, V in L
    4. Calculating the average kinetic energy (3/2)RT in J/mol
    5. P in torr, V in L
    6. P in Pa, V in m³
    7. P in atm, V given in mL
    8. The problem gives 101 kPa and 22.4 L

    Common mistakes

    Mistake 1: “Just plug in 25 for the temperature.” T in PV = nRT must be in kelvins. Using Celsius gives absurd results, like V = 0 at 0 °C.

    Mistake 2: “R is 8.314, whatever the units.” With pressure in atm, R is 0.08206 L·atm/(mol·K). Pairing atm with 8.314 makes the answer off by a factor of about 100. Check the units of P and V first, then choose R.

    Mistake 3: “PV = nRT is exact for every gas under all conditions.” It describes an ideal gas. At ordinary conditions the error is tiny, but at very high pressure or very low temperature real gases deviate noticeably.

    Quick check

    What is the volume of 2.00 mol of gas at 300 K and 1.00 atm?

    A gas has a density of 1.25 g/L at STP. What is its molar mass, approximately?

    Summary

    • The ideal gas law is PV = nRT, with R = 0.08206 L·atm/(mol·K) = 62.36 L·torr/(mol·K) = 8.314 L·kPa/(mol·K) = 8.314 J/(mol·K), chosen to match the units. T is in kelvins.
    • Since n = m/M, M = mRT/(PV) and d = PM/(RT): measuring a gas’s mass or density gives its molar mass.
    • An ideal gas has particles with no volume and no attractions. Real gases deviate at high pressure and low temperature.
    On to AP (for later)

    The AP equation sheet gives PV = nRT with R = 8.314 J/(mol·K) = 0.08206 L·atm/(mol·K) = 62.36 L·torr/(mol·K). AP labs often find a molar mass from data and then ask about error, for example whether water left in the flask makes the result too high or too low. Unit 3 also explains why real gases deviate from PV = nRT, using particle volume and intermolecular attractions.

    Data sources: the exact value of R is the CODATA 2018 recommended value.

    Practice this chapter: 2.8 · Gases · 10 questions

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