Gases - L9.2 The Ideal Gas Law

Ideal Gas Law

  • The ideal gas law describes the physical behavior of an ideal gas in terms of pressure, volume, temperature, and the number of moles of gas present.

PV=RTnPV=RTn

Practice Problems (Page 292)

Example Problems and Solutions

  • Problem 26: Determine the Celsius temperature of 2.49 mol of a gas contained in a 1.00-L vessel at a pressure of 143 kPa.

    • Solution:

      • Convert kPa to atm: 143kPa×1.00atm101.3kPa=1.41atm143 kPa \times \frac{1.00 atm}{101.3 kPa} = 1.41 atm

      • Use the ideal gas law to solve for T: T=PVnR=(1.41atm)(1.00L)(2.49mol)(0.0821L⋅atmmol⋅K)=6.90KT = \frac{PV}{nR} = \frac{(1.41 atm)(1.00 L)}{(2.49 mol)(0.0821 \frac{L \cdot atm}{mol \cdot K})} = 6.90 K

      • Convert Kelvin to Celsius: 6.90K−273=−266°C6.90 K - 273 = -266°C

  • Problem 27: Calculate the volume of a 0.323-mol sample of a gas at 265 K and 0.900 atm.

    • Solution: Use the ideal gas law to solve for V: V=nRTP=(0.323mol)(0.0821L⋅atmmol⋅K)(265K)0.900atm=7.81LV = \frac{nRT}{P} = \frac{(0.323 mol)(0.0821 \frac{L \cdot atm}{mol \cdot K})(265 K)}{0.900 atm} = 7.81 L

  • Problem 28: What is the pressure, in atmospheres, of a 0.108-mol sample of helium gas at a temperature of 20.0°C if its volume is 0.505 L?

    • Solution:

      • Convert Celsius to Kelvin: T=20.0°C+273=293KT = 20.0°C + 273 = 293 K

      • Use the ideal gas law to solve for P: P=nRTV=(0.108mol)(0.0821L⋅atmmol⋅K)(293K)0.505L=5.14atmP = \frac{nRT}{V} = \frac{(0.108 mol)(0.0821 \frac{L \cdot atm}{mol \cdot K})(293 K)}{0.505 L} = 5.14 atm

  • Problem 29: If the pressure exerted by a gas at 25°C in a volume of 0.044 L is 3.81 atm, how many moles of gas are present?

    • Solution:

      • Use the ideal gas law to solve for n: n=PVRT=(3.81atm)(0.044L)(0.0821L⋅atmmol⋅K)(298K)=6.9×10−3moln = \frac{PV}{RT} = \frac{(3.81 atm)(0.044 L)}{(0.0821 \frac{L \cdot atm}{mol \cdot K})(298 K)} = 6.9 \times 10^{-3} mol

  • Problem 30 (Challenge): An ideal gas has a volume of 3.0 L. If the number of moles of gas and the temperature are doubled while the pressure remains constant, what is the new volume?

    • Solution:

      • Since P and R are constants, they can be removed from the ideal gas law equation.

      • V1n1T1=V2n2T2\frac{V1}{n1T1} = \frac{V2}{n2T2} where n2=2n1n2 = 2n1 and T2=2T1T2 = 2T1

      • V1n1T1=V2(2n1)(2T1)⇒V11=V24⇒V2=4V1\frac{V1}{n1T1} = \frac{V2}{(2n1)(2T1)} \Rightarrow \frac{V1}{1} = \frac{V2}{4} \Rightarrow V2 = 4V1 V2=4(3.0L)=12LV_2 = 4(3.0 L) = 12 L

Ideal vs. Real Gases

  • Ideal gases obey all gas laws under all conditions of pressure and temperature.

  • Real gases behave like ideal gases and obey gas laws only at low pressure and high temperature, because the amount of empty space between the molecules increases and the forces of attraction are negligible.

Review: Avogadro's Principle & Ideal Gas Law

  • Avogadro's principle states that equal volumes of gases at the same pressure and temperature contain equal numbers of particles.

  • The ideal gas law relates the amount of a gas present to its pressure, temperature, and volume.

  • The ideal gas law can be used to find molar mass (if the mass of the gas is known) or the density of the gas (if its molar mass is known).

  • At very high pressures and very low temperatures, real gases behave differently than ideal gases.