Laws of Thermodynamics

Zeroth Law Of Thermodynamics And Temperature

If a system A is in thermal equilibrium with system B and the system B is in thermal equilibrium with system C, then systems A and C are in thermal equilibrium with each other.

Temperature scales

Relation between Celsius, Kelvin and Fahrenheit Scale:

$\frac{C}{5}=\frac{K-273}{5}=\frac{F-32}{9}$

Example 1: The electrical resistance of pure platinum increases linearly with increasing temperature. This property is used in a Platinum resistance thermometer. The relation between R_$\theta$ (Resistance at $\theta$K) and R₀ (Resistance at $\theta$₀K) is given as $R_{\theta }=R_o\left[1+\alpha \left(\theta -{\theta }_o\right)\right]$

where a = temperature coefficient of resistance. Now, if a Platinum resistance thermometer reads 0⁰ Celsius when its resistance is 80$\Omega$ and 100^o when its resistance is 90$\Omega$, find the temperature at which its resistance is 86$\Omega$.

Solution: Using the given relationship,

we have $\left(90-80\right)\Omega =\alpha \left(80\Omega \right)\left(100\right)$ . . . (i)

$\left(86-80\right)\Omega =\alpha [80\Omega ][\theta ]$. . . (ii)

Where $\theta$ is the desired temperature.

Taking the ratio of (i) & (ii)

First Law of Thermodynamics

First law of thermodynamics is simply a re-statement of the principle of conservation of energy for a thermally isolated system.

If Q, $\Delta$U & $\Delta$W represent the heat given to the system, change in its internal energy and the work done by the system respectively, the first law of thermodynamics states that,

$\Delta$Q = $\Delta$U + $\Delta$W

The heat transferred to the system ($\Delta$Q) is either utilised to do work ($\Delta$W) or increase the internal energy of the system ($\Delta$U).

Example 2: 3000 J of heat is given to a gas at constant pressure of 2 ´ 10⁵ N/m². If its volume increases by 10 litres during the process find the change in the internal energy of the gas

Solution: $\Delta$Q = 3000 J

W = P $\Delta$ V = (2´10⁵ N/m²) (10 x10^-3m³)

= 2 x 10³ J

$\Delta$U = $\Delta$Q – W = 3000 - 2000= 1000 J.

Second law of thermodynamics

(i) Kelvin Statement:- It is impossible to derive a continuous supply of work by cooling a body to a temperature lower than that of the coldest of its surroundings.

(ii) Clausius Statement:-It is impossible for a self acting machine, unaided by an external agency to transfer heat from a body to another at higher temperature.

Reversible Process:

A process which can be made to proceed in the reverse direction by variations in its conditions so that all changes occurring in any part of the direct process are exactly reversed in the corresponding part of the reverse process is called a reversible processes.

Irreversible Process:

A process which can not be made to proceed in the reverse direction is called an irreversible process.