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JEE Physics Thermodynamics Zeroth Law of Thermodynamics

Zeroth Law of Thermodynamics

In this article we will discuss Zeroth Law of Thermodynamics

In this article, we will discuss Zeroth Law of Thermodynamics. So, let’s get started.

Zeroth’s Law of Thermodynamics

If system A and system B are in thermal equilibrium with system C (i.e. they have the same temperature as C), they they are in thermal equilibrium with each other (i.e. they have the same temperature).

This property of nature enables one to assign a number value to temperature. We just need to choose two reference values, and use a substance with a macroscopic quantity that varies with ”our preception of temperature”. The substance we call a thermometer.

The Celsius choice of reference values is to assign the triple point of water (where H2O exists in solid, liquid and vapor forms) to be zero degrees. The other reference value is assigned to be 100 degrees where water and steam co-exist at one atmosphere pressure. Suppose we have a third substance at an unknown temperature T. How do we determine T? One way is to pick a substance to be a thermometer and define T to be proportional to a macroscopic quantity of the substance. Let’s demonstrate this with an example.

We will find the value of T of an object using three different substances for thermometers: liquid mercury, a resistor and a dilute gas. As our parameter we will choose the volume of mercury, the resistance of the resistor and the pressure of the dilute gas. Here is our (made-up) data:

• Temperature – Height of Hg – Resistance – Pressure of Dilute Gas

(1) 100 – 8.0 cm – 3.40 mV – 1.37 × 105 Pa
(2) T – 6.5 cm – 3.24 mV – 1.23 × 105 Pa
(3) 0 – 4.0 cm – 3.00 mV – 1.00 × 105 Pa

The ”unknown” temperature T of the object is determined for each thermometer by having the temperature be proportional to the parameter that is changing. So using
the mercury thermometer,
T = 100(6.5 − 4.0/8.0 − 4.0) = 62.5° (1)

Using the resistance thermometer,
T = 100(3.24 − 3.0/3.4 − 3.0) = 60.0°(2)

Using the dilute gas thermometer,
T = 100(1.23 − 1.0/1.37 − 1.0) = 62.2° (3)

The data was ”made-up”, however if real data were used the results would be sim￾ilar: in general the temperature T can depend on the type of thermometer used.

This is because temperature is defined to make the parameter of the thermometer proportional to T. All thermometers will agree at the two reference temperatures. However, in general, the parameters for each thermometer do not have exactly the
same behavior between the reference points.

Which substance is the best to use as a thermometer?

Is it possible to define temperature such that the equations of thermodynamics have a simple form
for all substances?

It would be nice if there were a class of substances that all give the same value for T. There is. After trying different types of thermometers we would discover the following property about the dilute gas thermometer above: we get the same value for T for any gas that we use! That is, if we use air, or H2, or O2, or He, etc. the value of T for our object, obtained by measuring the pressure of the gas, would be the same in each case. There are many proerties that all dilute gases have in common. Thus, if we quantify temperature using the constant volume dilute gas thermometer, there is a good chance that this method gives a useful definition of temperature. This will be our first way to quantify temperature

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