Thermodynamics In Physics : Thermodynamics is the study of the relations between heat, work, temperature, and energy. Thermodynamics is concerned with the work done by a system and the heat it exchanges with its surroundings. Alternatively, it is the study of changes that occur in some part of the universe (we designate as the system) and then everything else (outside the system) is the surrounding.
A real or imagined boundary may separate the system from its surroundings. A collection of properties such as pressure, volume, temperature and some other properties to be discussed later characterize the thermodynamical state of a system. The laws of thermodynamics describe how the energy in a system changes and whether the system can perform any useful work on its surroundings.
Thermodynamic Process : Thermodynamics basically deals with the exchange of heat between a body and the surrounding along with the other processes accompanying it, such as the work involved and the changes in the internal energy, taking place in the body. Such studies are primarily carried out in terms of a few macroscopic properties, such as pressure, volume, temperature, etc.
Any alteration in one or more of the macroscopic properties of a body renders specific changes in other related properties, and the body is said to have been subjected to a thermodynamic process (or is said to have undergone a thermodynamic process). The body which is subjected to a thermodynamic process is known as a thermodynamic system. The macroscopic parameters in terms of which the study of a thermodynamic process is carried out are known as thermodynamic variables. For example, gas enclosed in a cylinder fitted with a piston forms the thermodynamic system but the atmospheric air around the cylinder, movable piston, burner, etc., is the surrounding. The process of change of state of a system involves change of thermodynamic variables such as pressure P, volume V and temperature T of the system. During such a process, energy may be transferred into the system from the thermal reservoir (positive heat) or vice versa (negative heat). The process is known as thermodynamic process. Some important processes are: Isothermal process: Temperature remain constantQ.1. A gas is taken from state-1 to state-2 along the path shown in figure.
If 70 cal of heat is extracted from the gas in the process, calculate the change in internal energy of the system.
Ans. Since work done is equal to the area under the P-V graph, so in this case work done will be the negative shaded area as shown in Figure. Negative because volume is decreasing
Since, 70 cal of heat is extracted in the process, so
Q = –70 cal = 70 × 4.2 J = –294 J
From First Law of Thermodynamics (FLTD), we get
Hence, the internal energy of gas decreases by 241.5 J in the given process.
2. A sample of an ideal gas is taken through the cyclic process abca (figure ). It absorbs 50 J of heat during the part ab, no heat during bc and rejects 70 J of heat during ca. 40 J of work is done on the gas during the part bc. (a) Find the internal energy of the gas at b and c if it is 1500 J at a. (b) Calculate the work done by the gas during the part ca.
Ans. (a) In the part ab the volume remains constant. Thus, the work done by the gas is zero. The heat absorbed by the gas is 50 J. The increases in internal energy from a to b is
As the internal energy is 1500 J at a , it will be 1550 J at b . In the part bc , the work done by the gas is and no heat is given to the system. The increase n internal energy from b to c is
As the internal energy is 1550 J at b , it will be 1590 J at c .
(b) The change in internal energy from c to a is
The heat given to the system is
Using
3. Heat taken from a gas in a process is 80 J and increase in internal energy of the gas is 20 J. Find work done by the gas in the given process.
Ans. Heat is taken from the gas. Therefore, Q is negative. Or, Q = –80 J
Internal energy of the gas is increasing
Therefore, is positive. Or
Using the first law equation,
Here, negative sign indicates that volume of the gas is decreasing and work is done on the gas.
4. The temperature of n -moles of an ideal gas is increased from T 0 to 2 T 0 through a process Find work done in this process.
Ans. pV = nRT (ideal gas equation) …(i)
And …(ii)
Dividing Eq. (i) by Eq. (ii), we get
= 2 nRT 0