Which law states that the pressure and absolute temperature of a fixed quantity of gas are directly proportional under constant volume conditions?
Answers
Charles' Law – Relationship between Volume and Temperature
Charles and Gay-Lussac each studied the effect of temperature on a sample of gas at a constant pressure.
Gas was placed in a tube with a mercury plug over it.
This provided a constant external pressure.
As the gas was heated, the volume of the gas increased. As the gas was cooled, the volume decreased.
Lord Kelvin used this same experiment to calculate the value of absolute zero, -273.15 oC. At this temperature, the volume of an ideal gas goes to zero.
Charles' Law states that a constant pressure, the volume occupied by a fixed amount of gas is directly proportional to its absolute temperature(K).
V ∝T
The volume of a fixed amount of gas in a container at constant pressure is directly proportional to the absolute temperature (K).
Charles' Law can be used to solve certain types of problems, those in which two of the gas variables, n (moles of gas) and V (volume) remain constant. There can be a change in either T or P.
Using Charles' Law we can predict how this change will affect the other factor. The temperature must always be expressed in K.
425 mL of Ar gas are heated from 22oC to 187oC, at constant pressure. What is the final volume of the gas?
initial conditions final conditions
T1 = 22oC T2 = 187oC
V1 = 425 mL V2 =
V/T = constant so V1/T1 = V2/T2
The temperature must be expressed in K.
T1 = 273.2 + 22 = 295.2 K
T2 = 273.2 + 187 = 460.2 K
A container is filled with a sample of N2 gas at 1 atm pressure. It is immersed in an ice water bath at 0oC. The volume of the gas at this temperature is found to be 58.7 mL. The container is then placed in a bath of boiling isopropyl (rubbing) alcohol. The volume of the gas is found to be 76.4 mL. What is the boiling point of isopropyl alcohol? The pressure of the gas remains constant at 1 atm.
initial conditions final conditions
T1 = 0oC T2 =
V1 = 58.7 mL V2 = 76.4 mL
V/T = constant so V1/T1 = V2/T2
Again, the temperature must be expressed in K.
T1 = 273.2 + 0 = 273.2 K
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