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The ideal gas law connects pressure, volume, amount of gas, and temperature in one useful equation. This cheat sheet helps students set up and solve common gas law problems using PV=nRTPV = nRT. It is especially useful when problems include unit conversions or ask for an unknown variable.

Clear walkthrough habits reduce algebra mistakes and help students choose the correct value of RR.

The core idea is that gas behavior depends on four variables: pressure PP, volume VV, moles nn, and Kelvin temperature TT. The ideal gas law is PV=nRTPV = nRT, and it can be rearranged to solve for any one unknown. Temperature must always be in Kelvin, so use K=C+273.15K = ^{\circ}C + 273.15.

Units must match the gas constant, such as R=0.0821 LatmmolKR = 0.0821\ \frac{L\cdot atm}{mol\cdot K} when pressure is in atmospheres and volume is in liters.

Key Facts

  • The ideal gas law is PV=nRTPV = nRT, where PP is pressure, VV is volume, nn is moles, RR is the gas constant, and TT is temperature in Kelvin.
  • To solve for pressure, rearrange the equation as P=nRTVP = \frac{nRT}{V}.
  • To solve for volume, rearrange the equation as V=nRTPV = \frac{nRT}{P}.
  • To solve for moles, rearrange the equation as n=PVRTn = \frac{PV}{RT}.
  • To solve for temperature, rearrange the equation as T=PVnRT = \frac{PV}{nR}.
  • Use R=0.0821 LatmmolKR = 0.0821\ \frac{L\cdot atm}{mol\cdot K} when pressure is measured in atmatm and volume is measured in LL.
  • Convert Celsius to Kelvin with K=C+273.15K = ^{\circ}C + 273.15 before using any gas law calculation.
  • At standard temperature and pressure, one mole of an ideal gas has a volume of about 22.4 L22.4\ L at 273.15 K273.15\ K and 1 atm1\ atm.

Vocabulary

Ideal gas law
The equation PV=nRTPV = nRT that relates pressure, volume, moles, and Kelvin temperature for an ideal gas.
Pressure
Pressure is the force of gas particle collisions per unit area, often measured in atmatm, kPakPa, or mmHgmmHg.
Volume
Volume is the amount of space a gas occupies, commonly measured in liters LL for ideal gas law problems.
Mole
A mole is an amount of substance equal to 6.022×10236.022 \times 10^{23} particles.
Kelvin
Kelvin is the absolute temperature scale used in gas law calculations, found with K=C+273.15K = ^{\circ}C + 273.15.
Gas constant
The gas constant RR is the proportionality constant in PV=nRTPV = nRT, and its value depends on the pressure and volume units used.

Common Mistakes to Avoid

  • Using Celsius directly is wrong because gas law equations require absolute temperature. Always convert with K=C+273.15K = ^{\circ}C + 273.15 before substituting into PV=nRTPV = nRT.
  • Mixing units with the wrong gas constant is wrong because RR must match the pressure and volume units. If using R=0.0821 LatmmolKR = 0.0821\ \frac{L\cdot atm}{mol\cdot K}, pressure must be in atmatm and volume must be in LL.
  • Forgetting to rearrange the equation before substituting can lead to algebra errors. For example, when solving for moles, use n=PVRTn = \frac{PV}{RT} instead of trying to divide randomly after plugging in numbers.
  • Rounding too early can make the final answer noticeably inaccurate. Keep several digits during the calculation and round the final answer to the correct number of significant figures.
  • Ignoring units in the setup is wrong because units show whether the equation is being used correctly. A correct setup should allow units such as atmatm, LL, molmol, and KK to cancel or match properly.

Practice Questions

  1. 1 A gas sample has P=1.25 atmP = 1.25\ atm, V=3.40 LV = 3.40\ L, and T=298 KT = 298\ K. How many moles of gas are present using R=0.0821 LatmmolKR = 0.0821\ \frac{L\cdot atm}{mol\cdot K}?
  2. 2 What volume will 0.750 mol0.750\ mol of gas occupy at 2.00 atm2.00\ atm and 315 K315\ K using R=0.0821 LatmmolKR = 0.0821\ \frac{L\cdot atm}{mol\cdot K}?
  3. 3 A container holds 1.50 mol1.50\ mol of gas in 10.0 L10.0\ L at 25.0C25.0^{\circ}C. What is the pressure in atmatm after converting temperature to Kelvin?
  4. 4 Why must temperature be converted to Kelvin before using PV=nRTPV = nRT, and what could happen if 25C25^{\circ}C were used directly instead of 298.15 K298.15\ K?