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Kelvin scale of temperature
Temperature is a measure of how hot or cold a substance is. Temperature is often measured in degrees Celsius (oC) but in physics we use the unit kelvin (K).The kelvin temperature of a gas is proportional to the average kinetic energy of its molecules.
The higher the temperature in kelvins, the more kinetic energy the gas molecules have and the faster they move. The kelvin scale of temperature differs to the Celsius scale as it starts from 0 and includes no negative numbers. The minimum value on the kelvin scale is 0 K. This is also known as absolute zero and is equal to -273oC on the Celsius scale.
At absolute zero the particles have no energy whatsoever and do not move at all. Particles cannot have negative movement so the scale cannot possibly go below absolute zero (0 K).
The kelvin scale starts at 0 K and can go up to large numbers. You should be able to convert temperatures between kelvins and degrees Celsius.
- To convert a temperature from Celsius to kelvin, you add 273
- To convert a temperature from kelvin to Celsius, you subtract 273
- An increase in temperature of 1 kelvin is equivalent to an increase of 1 Celsius
Example
1. Convert the following temperatures from Celsius to kelvin.
a) 0oC
b) 100oC
c) 50oC
2. Convert each of the following temperatures from kelvin to Celsius.
a) 0 K
b) 100 K
c) 400 K
| 1. a) 0oC + 273 = 273 Kb) 100oC + 273 = 373 Kc) 50oC + 273 = 323 K | 2. a) 0 K – 273 = -273oCb) 100 K – 273 = -173oC c) 400 K – 273 = 127oC |
Relating temperature and pressure
Robert Boyle carried on with his investigations into the behaviour of gases. Boyle knew that there was a relationship between the temperature and pressure of a gas and conducted a series of experiments to find out what this relationship was.
He measured the pressure of gases at various different temperatures and found that the lower the temperature, the lower the pressure exerted by the gas, and the higher the temperature, the higher the pressure exerted. This qualitative relationship can be explained by looking at the link between kinetic energy and temperature.
At 0 K the particles have zero energy and do not move. As the temperature is increased above absolute zero, the particles gain kinetic energy and move faster. Therefore, an increase in the temperature in kelvins causes the particles to move faster and faster. The faster the particles move, the more they collide with one another and the walls of any container they are enclosed in. This increase in the number of collisions results in an increase in pressure exerted by the gas.
In the graph above we can see that an increase in temperature (along the axis) results in an increase in pressure (along the axis). From his experiments, Boyle discovered that the kelvin temperature of a gas is proportional to the average kinetic energy of its molecules.
This relationship is shown by the equation:
![\[ \frac{p_1}{T_1} = \frac{p_2}{T_2} \]](https://online-learning-college.com/wp-content/ql-cache/quicklatex.com-779519e51f68c5b8334a1ee0c0895702_l3.png "Rendered by QuickLaTeX.com")
p1 is the initial pressure of the gas and T1 is the initial temperature.
p2 is the final pressure of the gas and T2 is the final temperature.
For this equation, temperature must always be measured in kelvin (K) and pressure is measured in pascals (Pa).
The pressure and temperature of a gas could be measured and used in this equation to calculate the pressure of the same gas at any other temperature.
Example
The pressure of Gas A is found to be 100 kPa at a temperature if 20 The same gas was then heated to a temperature of 60. Calculate the new pressure of the gas at this higher temperature.
We are provided with the initial temperature and pressure of the gas as well as the final temperature of the gas. However, the temperatures are quoted in degrees Celsius so must be converted into kelvin. To do this we need to add 273 to both values:
![\[ 20\degree C = 293 K \]](https://online-learning-college.com/wp-content/ql-cache/quicklatex.com-4b4b8cb37b26697f296191a83a5aac8c_l3.png "Rendered by QuickLaTeX.com")
![\[ 60\degree C = 333 K \]](https://online-learning-college.com/wp-content/ql-cache/quicklatex.com-a97317695a81781c534e1073b15d38c3_l3.png "Rendered by QuickLaTeX.com")
We now know that p1 = 100Pa, T1 = 293K and T2 = 333K and and we can substitute all values into the equation:
![\[ \frac{100 Pa}{293 K} = \frac{p_2}{333 K} \]](https://online-learning-college.com/wp-content/ql-cache/quicklatex.com-750fa1a6fe686a8a4f45a9642545c5b5_l3.png "Rendered by QuickLaTeX.com")
![\[ \frac{100 Pa}{293 K} \times 333K = p_2 \]](https://online-learning-college.com/wp-content/ql-cache/quicklatex.com-f6dc7344f6dc071c3d40f44b5640f1d4_l3.png "Rendered by QuickLaTeX.com")
![\[ p_2 = 113.65 Pa \]](https://online-learning-college.com/wp-content/ql-cache/quicklatex.com-af808cc065ca4c1a223af08d83f2e10b_l3.png "Rendered by QuickLaTeX.com")
An increase in temperature of 40 K caused the pressure to increase by 13.65 Pa.
