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The ideal gas law connects pressure, volume, temperature, and amount of gas in one powerful equation. Students need this cheat sheet because many gas law problems are not hard once the units and substitutions are organized correctly. Worked examples help show when to solve for moles, pressure, volume, temperature, molar mass, or density. This reference is especially useful for chemistry and physics problems involving gases in containers, cylinders, and laboratory conditions. The core equation is PV=nRTPV = nRT, where PP is pressure, VV is volume, nn is moles, RR is the gas constant, and TT is absolute temperature. Temperature must be in kelvin using TK=TC+273.15T_K = T_C + 273.15, and the value of RR must match the pressure and volume units. Many examples also use n=mMn = \frac{m}{M} to connect mass and molar mass, or ρ=PMRT\rho = \frac{PM}{RT} to connect gas density to molar mass. Careful unit matching and algebra are the main skills needed for success.

Key Facts

  • The ideal gas law is PV=nRTPV = nRT, and it relates pressure, volume, moles, and absolute temperature for an ideal gas.
  • Temperature must be converted to kelvin before using the ideal gas law, so TK=TC+273.15T_K = T_C + 273.15.
  • Use R=0.08206 LatmmolKR = 0.08206\ \frac{\text{L}\cdot\text{atm}}{\text{mol}\cdot\text{K}} when pressure is in atmospheres and volume is in liters.
  • Use R=8.314 JmolKR = 8.314\ \frac{\text{J}}{\text{mol}\cdot\text{K}} when pressure is in pascals and volume is in cubic meters, since 1 J=1 Pam31\ \text{J} = 1\ \text{Pa}\cdot\text{m}^3.
  • To solve for moles, rearrange the ideal gas law as n=PVRTn = \frac{PV}{RT}.
  • To connect mass to gas amount, use n=mMn = \frac{m}{M}, where mm is mass and MM is molar mass.
  • Gas density can be found from the ideal gas law using ρ=PMRT\rho = \frac{PM}{RT}.
  • At standard temperature and pressure for many classroom problems, T=273.15 KT = 273.15\ \text{K}, P=1 atmP = 1\ \text{atm}, and one mole of ideal gas occupies about 22.4 L22.4\ \text{L}.

Vocabulary

Ideal gas
An ideal gas is a model gas whose particles have negligible volume and no intermolecular attractions.
Pressure
Pressure is the force per unit area caused by gas particles colliding with the walls of a container.
Absolute temperature
Absolute temperature is temperature measured in kelvin, where 0 K0\ \text{K} represents absolute zero.
Mole
A mole is an amount of substance equal to 6.022×10236.022 \times 10^{23} particles.
Gas constant
The gas constant RR is the proportionality constant in PV=nRTPV = nRT and its numerical value depends on the units used.
Molar mass
Molar mass MM is the mass of one mole of a substance, usually measured in g/mol\text{g/mol}.

Common Mistakes to Avoid

  • Using Celsius directly in PV=nRTPV = nRT is wrong because the ideal gas law requires absolute temperature in kelvin. Always convert with TK=TC+273.15T_K = T_C + 273.15 before substituting.
  • Mixing unit systems is wrong because the gas constant RR only works with specific pressure and volume units. Use R=0.08206 LatmmolKR = 0.08206\ \frac{\text{L}\cdot\text{atm}}{\text{mol}\cdot\text{K}} with liters and atmospheres, or R=8.314 JmolKR = 8.314\ \frac{\text{J}}{\text{mol}\cdot\text{K}} with pascals and cubic meters.
  • Forgetting to convert mass to moles is wrong because PV=nRTPV = nRT uses nn, not mass directly. Use n=mMn = \frac{m}{M} before using the ideal gas law.
  • Solving algebra after substituting messy numbers can lead to errors because the unknown may be buried in the equation. Rearrange first, such as V=nRTPV = \frac{nRT}{P} or P=nRTVP = \frac{nRT}{V}, then substitute units.
  • Assuming every real gas behaves ideally is wrong because high pressure and low temperature can make particle volume and attractions important. The ideal gas law works best at low pressure and high temperature.

Practice Questions

  1. 1 A gas sample has n=0.750 moln = 0.750\ \text{mol}, T=298 KT = 298\ \text{K}, and P=1.20 atmP = 1.20\ \text{atm}. Find the volume in liters using R=0.08206 LatmmolKR = 0.08206\ \frac{\text{L}\cdot\text{atm}}{\text{mol}\cdot\text{K}}.
  2. 2 A rigid container has volume V=2.50 LV = 2.50\ \text{L} and contains 0.150 mol0.150\ \text{mol} of gas at 35.0C35.0^\circ\text{C}. Find the pressure in atmospheres.
  3. 3 A 4.00 g4.00\ \text{g} sample of oxygen gas, O2\text{O}_2, is held at T=300 KT = 300\ \text{K} and P=0.950 atmP = 0.950\ \text{atm}. Use M=32.0 g/molM = 32.0\ \text{g/mol} to find the volume.
  4. 4 A sealed rigid tank is heated while the amount of gas stays constant. Explain why the pressure increases using the relationship PV=nRTPV = nRT and the particle collision model.