# Charles's Law

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Introduction

## Charles's Law

## The pressure that a gas exerts on the walls of its container is determined by the momentum of the atoms and the molecules of the gas, which in turn is determined by temperature. As the temperature increases the atoms and molecules move faster, and so exert a greater pressure on the walls. If the walls are rigid, such that the volume of the container is held constant, the relationship between pressure P and temperature T is given by P = constant x T.

## However, if the walls are flexible, as the temperature increases the volume increases to maintain even pressure. This is called Charles’s Law.

Charles law states that the volume of a given amount of dry ideal gas is directly propotional to the Kelvin Temperature, provided the amount of gas and the pressure remains fixed.i.e.

V=constant ( t + 273.15)Where t is the gas temperature on Celsius scale.

There are interesting points regarding the relationship between volume and temperature changes.

-273ºCT (ºC)

Since gases expand and contract at a constant rate, extrapolation of this behavior shows that the effective volume goes to zero around -273 ºC.

Middle

or

=const.

T

or

## =

T1T2

Charles's law is a special case of the perfect gas equation andconcludes that the volume of any gas should extrapolate to zero at t = - 273.15 C. This law may be stated as following equation:

V=constant T where T is the absolute temperature.

## The aim of this investigation is to prove Charles’s law and through that to find a value for absolute zero. To prove this law, I shall carry out an experiment in which a volume of air is trapped inside a capillary tube at a constant pressure, this is attached to a thermometer and placed in water at various temperatures. According to Charles’s theory the gas will expand, and by measuring the diameter of the tube and the height of the column of air, the volume can be calculated (over a temperature range of 100°C).

## Plan/method

## Variables -

## diagram

Conclusion

Rearranged as: a = F

m

showing that acceleration is indirectly proportional to the mass if the force releasing the ball bearing remains constant. Therefore, if the mass is increased, the acceleration of the ball bearing will be decreased during its trajectory.

Theory for horizontal motion:

Loss of PE = Gain of PE

mgh = ½ mv2

2mgh = v2

m

v = Ö(2gh)

It is range we are trying to find out:

s = ½ (u+v) t = ut = vt

R = Ö(2gh) = Ö(2H/g) (from equation: ‘time of flight´)

R = Ö(4hH)

R2 = 4hH

From this derived equation a testable hypothesis can now be made. By comparing the derived formula with the formula for a straight line graph, which shows similar properties, it is thought that as height increases, range will also increase.

R2 = 4 x h x H

R2 = (4 x h) x H + 0

y = m x + c

This comparison of equations shows that if Range2 were plotted against height (h) then a straight line graph through the origin would be the result. It can therefore be said that Range2 is directly proportional to height, considering that height (H) is constant. The mass and size of the ball bearing must also be kept constant. To prove this, a graph of Range2 on the y-axis and height (h) on the x-axis must be plotted and a straight line graph running through the origin must be produced from the data. This would show the theory and equations used to be correct.

This student written piece of work is one of many that can be found in our AS and A Level Mechanics & Radioactivity section.

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