Boyle’s Law - Review P1V1 = P2V2
Pressure can be in any units Volume can be in any units Boyle’s Law P1V1 = P2V2 Pressure can be in any units Volume can be in any units
Charles’ Law If temperature increases, volume increases If temperature decreases, volume decreases This is direct variation (compared to inverse for Boyle’s law)
Charles Law Volume can be in any units Temperature must be in Kelvin V1 = V2 T1 T2 Volume can be in any units Temperature must be in Kelvin To convert Celsius to Kelvin, add 273 (K = C + 273.)
Charles law For example: The temperature inside my fridge is 4 C. If I place a balloon in my fridge that initially has a temperature of 22 C and a volume of .5 L, what will be the volume of the balloon when it is fully cooled by my refrigerator?
Charles’ Law First convert temperatures to Kelvin 4 + 273 = 277K
Charles’ Law Make a list of what you know V1 = T1 = V2 = T2 =
Charles’ Law V1 = .5 L T1 = V2 = T2 =
Charles’ Law V1 = .5 L T1 = 295K V2 = T2 =
Charles’ Law V1 = .5 L T1 = 295K V2 = x T2 =
Charles’ Law V1 = .5 L T1 = 295K V2 = x T2 = 277K
Charles’ Law Put the information into the equation: V1 = V2 T1 T2 .5 = x 295 277
Charles’ Law Cross multiply to solve for x .5(277) = x(295) x = .47 Liters
Charles’ Law For Example: A man heats a balloon in the oven. If the balloon initially has a volume of 0.4 L and a temperature of 20 C, what will the volume of the balloon be after he heats it to a temperature of 250 C?
Charles’ Law Convert temperatures to Kelvin 20 + 273 = 293 K
Charles’ Law Make a list of what you know V1 = T1 = V2 = T2 =
Charles’ law Make a list of what you know V1 = .4 L T1 = 293 K V2 = x
Charles’ law Put the information into the equation: V1 = V2 T1 T2 .4 = x 293 523
Charles’ law Cross multiply to solve for x .4(523) = x(293) x = .71 L