11.31 Gas Laws (Combined Reteach) Notes

STP
temperature of 0.00°C or 273 K and a pressure of 1.00 atm.

The combination of Boyle's Law and Charles's Law is called the Combined Gas Law

​Temperatures in ^oC +273.15 = K
ALWAYS check that your units match!
If P₁ is in atm P₂ also needs to be in atm

Key step Identifying Variables Combined Gas Law

• A helium balloon has a volume of 48.7 L at 37°C and 700 Pa . What Volume will it have at 970 Pa and 87^oC?

  • Step one Identify all the variables

    • V₁=

    • P₁=

    • T₁=

    • V₂=

    • P₂=

    • T₂=​

    ​

Convert ^oC!!!
C + 273.15 = K

Combined Gas Law

• A helium balloon has a volume of 48.7 L at 37°C and 700 Pa . What Volume will it have at 970 Pa and 87^oC?

  • Step one Identify all the variables

    • V₁= 48.7 L

    • P₁= 700 Pa

    • T₁= 37 ^oC +273.15 = 310.15 K

    • V₂= X

    • P₂= 970 Pa

    • T₂=​ 87 ^oC +​ 273.15 = 360.15 K

​

• V₁= 48.7 L

• P₁= 700 Pa

• T₁= 37 ^oC +273.15 = 310.15 K

• V₂= X

• P₂= 970 Pa

• T₂=​ 87 ^oC +​ 273.15 = 360.15 K

Combined Gas Law

• The volume of a gas at 45°C and 0.678 atm is 180 mL. What Volume will it have at standard Temperature and a new volume of 79 mL?

  • Step one Identify all the information we have

    • V₁=

    • P₁=

    • T₁=

    • V₂=

    • P₂=

    • T₂=​

    ​

Combined Gas Law

• The volume of a gas at 45°C and 0.678 atm is 180 mL. What Pressure will it have at standard Temperature and a new volume of 79 mL?

  • Step one: Identify all the information we have

    • V₁= 180 mL

    • P₁= 0.678 atm

    • T₁= 45 C + 273.15 = 318.15 K

    • V₂= 79 mL

    • P₂=

    • T₂=​ 273 K

    ​

​

• V₁= 180 mL

• P₁= 0.678 atm

• T₁= 45 C + 273.15 = 318.15 K

• V₂= 79 mL

• P₂=

• T₂=​ 273 K