Charles formulated this law in 1787 giving the relationship between volume and temperature of a gas. This law stated that at constant pressure, the volume of a given mass of gas increases or decreases by 1/273 of its volume at 0°C for every one degree rise or fall.
For one degree rise in temperature = V0 x (1/273)
The Volume Vt at any temperature‘t’ is given as V0 + {(V0xt)/(273)}
Or
Vt = V0 {1+(1/273)}
A plot of volume along x-axis and temperature along y-axis gives a straight line intercepting the y-axis. When this line is extrapolated to lower temperature, it cuts the x-axis, which represents the zero volume.
Fig: - Plot of volume versus temperature (P constant)
The temperature at which the volume of the gas becomes zero is found to be -273°C, which is independent of the nature and pressure of the gas. The lowest temperature below which volume does not exist (negative), is called the absolute zero. Temperature measurements based on the absolute zero is known as absolute scale of temperature or Kelvin temperature scale.
Charles' law is mathematically represented as:
vt = v0{1+(t/173)} = v0{(273+t)/273}
273 + t = T (K) and 273 =T0 which corresponds to 0°C on absolute temperature scale
vt = (v0T/T0) or (vt/v0) = (T/T0) = (vt/T) = (v0/T0)
Thus, V/T constant = k'
The above relation may be written as: V = k’T = V a T.
The volume of a given mass of a gas is directly proportional to the absolute temperature at constant pressure. If V1 is the volume of a certain mass of gas at temperature T1 and V2 is the volume of the same mass of the same gas at temperature T2 at constant pressure, then,
V1 / T1 = V2 / T2
The validity of Charles' law can be determined by measuring the volumes of a given mass of a gas at different temperatures, at constant pressure. The value of V/T remains to be constant in this study. The curves obtained by plotting volume temperature ratio against different pressures are called isobars. Charles described the expansive nature of gases with lower density.
Problem
2. A sample of helium has a volume of 520 cm3 at 100°C. Calculate the temperature at which the volume will become 26cm3. Assume that pressure is constant?
Solution
V1 = 520 cm3 V2 = 260 cm3
T1 = 100 + 273 = 373K T2 = ?
V1 / T1 = V2 / T2
Or
T2 = ((V2 x T1) / V1)
T2 = 186.5 K
t = 186.5 - 273 = 86.5°C.
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