or
=
T1 T2
Charles's law is a special case of the 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
How did the did the volume- temperature data obtained by Charles lead to the concept of absolute zero?
Question: How did the did the volume- temperature data obtained by Charles lead to the concept of absolute zero?
Answer:
Charles saw a linear relationship between the volume and temperature of a gas. Extrapolating backwards, he found that the point where a gas would have no volume would be -273 degrees Celsius. Since that's as cold as he thought things could ever get, that originated the idea of absolute zero.
As it turns out, 0 K is the lowest temperature that can be achieved, but not because of Charles law. It actually has to do with the fact that at absolute zero, molecules have the smallest amount of energy possible, and have ceased moving back and forth entirely. It's sometimes thought that molecules don't move at all at absolute zero - but there's still some vibrational motion in bonds even at this temp.
Relation between Temperature and Volume
When the experiment is conducted, it must be insured that all of the apparatus is attached securely to ensure that nothing comes apart. General laboratory rules must be recognised to ensure safety throughout.
The apparatus that is required is:
Curtain rail (ski slope) – reaching 0.50m in vertical height
Ball bearing
Metre ruler – 1m long measuring in 0.001m intervals
Carbon paper
Banner of white paper – approx. 3m long
Clamps
Retort Stands
Set square – measuring 90º angles
Horizontal bench – vertical height of between 0.8m and 1.0m is practical
The variable that will be investigated is the range that the ball bearing is projected.
The independent variable is the height up the ski slope that the ball is released from. It is measured using a metre ruler.
The dependant variable is the distance from the slope in the horizontal direction that the ball will travel.
The method that will be undertaken is firstly to set up the apparatus as shown in the diagram. The height above the ground, H, should be measured using a metre ruler and this should be kept constant throughout the experiment. The angle at which the ski slope is at should also be kept constant and should be held in place by clamps, so that the edge of the slope is parallel with the edge of the table. The banner of paper should be placed on the floor running away from the table. Practice runs of placing the ball at different heights should be done to find roughly the place where the carbon paper should be placed.
For the experiment, the ball bearing should be held in place on the ramp at certain vertical heights above the table. It can then be released and as the ball strikes the carbon paper it should leave a mark on the paper below. The length/range can then be measured from the edge of the table to the centre of the mark left by the carbon paper. The experiment can then be repeated 3 times for each height to reduce the effect of anomalous results.
The experiment willl be conducted using heights of release of the ball bearing from 0.00m to 0.40m in stages of 0.05m. This will hopefully obtain a good set of results.
In order to simplify this experiment, air resistance has been ignored as a factor in the range of the projectile. As it is a projectile that will be measured, the only other factor working on it are gravitational forces. The horizontal motion of the projectile is independent of the vertical motion, and that is why the falling motion of the ball bearing does not need to be considered.
A ball bearing will be used, as it is a fairly dense object. This has benefits because although we are ignoring the effect of air resistance in the calculations, we should minimise the affect that it has. A lighter object, or one that has a larger surface area (i.e. is less dense), would show more affect from air resistance. However, the smooth and relatively dense ball bearing should show little difference in trajectory due to the affects of air resistance whereas an object such as a polystyrene ball would be massively altered. This is known from the equation: F = ma
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.