Equation, Application, Limitation & Examples
First Law of thermodynamics
First Law of thermodynamics: There are four laws of thermodynamics, namely, Zeroth’s law of thermodynamics, the First law of thermodynamics, the Second Law of thermodynamics and the Third law of thermodynamics. Among the four laws, the first law works on the principle of the Law of conservation of energy. As we know, the internal energy (U) of a system can change through two energy transfer modes: heat and work. The heat supplied(Q) to the system goes on to increase the internal energy(U) and the rest goes into work done(W) on the environment. Whenever the system undergoes any change because of the interaction of heat, energy, and work, it follows the numerous energy transfer or conversions, but there is no net change in the total energy. Also, the first law of thermodynamics is sometimes also referred to as the Law of Conservation of energy.
According to the First law of thermodynamics, “ The heat given to the system by the surrounding is used to change the internal energy while the remaining is used in doing work by the system”.
First Law of Thermodynamics Equation
The first law of thermodynamics can be mathematically written as-
ΔQ = ΔU + W …………………. (1)
Where,
ΔQ is the heat supplied
ΔW is the work done
ΔU is the change in internal energy
Alternatively, Equation 1 can be written as-
ΔU = ΔQ – W …………………… (2)
Sign Convention in Relation to First Law of Thermodynamics
The sign conventions in relation to the first law of thermodynamics are-
| Sign Convention in Relation to First Law of Thermodynamics | |
| ΔU (change in internal energy) | “+” indicate temperature increases |
| “-” indicate temperature decreases | |
| “0” indicate constant temperature | |
| Q (heat) | “+” indicate heat enters gas |
| “-” indicate heat exits the gas | |
| “0” indicate no heat exchange | |
| ΔW (work done on the gas) | “+” indicate gas is compressed |
| “-” indicate gas expansion | |
| “0” indicate constant volume | |
Limitation of the First Law of Thermodynamics
The Limitation of the first law of thermodynamics is-
- The process under the first law of thermodynamics proceeds in a particular direction only. For example, Heat transfer takes place from the hot body to the cold body only. But the heat can not flow naturally from a cold body to a hot body.
- The first law only provides necessary but not sufficient conditions for a process to occur.
- All processes involving the conversion of heat into work and vice-versa are not equivalent. The work cannot be fully converted into heat, but a conversion of heat completely into the work is not possible.
Applications of the First Law of Thermodynamics
Some of the applications related to the first law of thermodynamics are-
Heat Engine: The application based upon the First Law is the heat engine. It converts thermal energy into mechanical energy and vice versa.
Refrigerators and Heat Pumps: Refrigerators and heat pumps are heat engines that convert mechanical energy to heat. It is just reverse to heat engine.
Solved Example of First Law of Thermodynamics
Example 1: Calculate the change in the system’s internal energy if 5000 J of heat is added to a system and a work of 3500 J is done.
Solution: According to the question, the heat (q) is 5000 J
Work done(w) is 3500 J
As we know that as the heat is added to the system, then according to sign convention it will remain positive.
According to the first law of thermodynamics, ΔQ = ΔU + W
Substituting the value, we get,
5000 = ΔU + 3500
ΔU = 5000 – 3500
ΔU = 1500 J
So the change in the Internal energy of the system is 1500 J.
Example 2: Complete the value of the table given below( All values given are in KJ)
| Q | W | U(initial) | U(final) | ΔU |
| ? | 6 KJ | ? | 45 | 34 |
Solution: From the table above we can say-
W = 6 KJ
U(final) = 45
ΔU = 34
U(initial) = ?
Q =?
According to the first law of thermodynamics, Q = ΔU + W
Q = 34 -6
Q = 28KJ
According to the first law of thermodynamics, Q = ΔU + W
We can say it as, Q = [U(final)- U(initial)] + W
28 = [45 – U(initial)] + 6
28 = 51 – U(initial)
U(initial) = 51 -28
U(initial) = 23 KJ
Example3: What is the change in the internal energy of the system if 4000 J of heat leaves the system and 4500 J of work is done on the system?
Solution: According to the question, the Heat(Q) of the system is 4000 J
Work done(W) is 4500 J, As the heat leaves the system then the work done will be negative as -4500J
According to the first law of thermodynamics, ΔQ = ΔU + W
We can say it as, ΔU = Q-W
ΔU = -4000-(-4500)
ΔU = -4000+4500
ΔU = 500 joules