First Law Of Thermodynamics

In the first law of thermodynamics, we make a relation between heat, work and internal energy. When a system experiences transformation influenced by the exchange of heat, work, and internal energy, it follows numerous energy transfers and conversions. But during these transfers, there is no net change in the total energy. In thermodynamic processes, the first law of thermodynamics follows the conservation of energy. Just like mass, energy is always conserved, i.e., it can neither be created nor destroyed; it can be transformed from one form to another form. 

1.0Statement and Equation of the First Law of Thermodynamics 

In Physics, The first law of thermodynamics essentially says that when you give heat to a system, the total heat the system takes in is used for two things: increasing its internal energy and doing external work. In simpler terms, the heat you put in gets split between making the system's inside energy go up and doing work outside the system. 

dQ = dU + dW OR  Q = ΔU + W

2.0Sign Convention used in Physics for First Law of Thermodynamics

FLOT, ΔQ = dU + ΔW or [Q = ΔU + W]

Sign convention:

3.0Important Points of the First Law of Thermodynamics 

• This law is applicable to every process in nature.

• Thermodynamic force is a non-conservative force.

• The first law of thermodynamics brings in the idea of internal energy. This is essentially the energy stored within a system, and it plays a crucial role in how heat is managed and utilized within that system.
• The first law of thermodynamics is based on the law of conservation of energy.
• Make sure that dQ, dU, and dW share the same units for a seamless equation. (either in units of work or in units of heat).

• This law holds true for all three states of matter: solid, liquid, and gas.

• dU is a characteristic of the state of a system; it may be any type of internal energy–translational kinetic energy, vibrational, rotational kinetic energy, binding energy, etc.

4.0Examples of the First Law of Thermodynamics

Example:

During an experiment heat released by the system is 10 cal and work done on the system is 10 J. If initial internal energy of the system is 50 J then find final internal energy.

Solution:

As ΔQ = ΔW + U_f – U_i

– 10 cal = – 10 J + U_f – 50

 ⇒ U_f = – 10 × 4.2 + 10 + 50

 ⇒ U_f = – 42 + 60 = 18 J

Example:

A gas under constant pressure of 4.5×10⁵ Pa when subjected to 800 kJ of heat, changes the volume from 0.5m³ to 2.0 m³. Find the change in internal energy of the gas?

Solution:

ΔU = ΔQ – ΔW = ΔQ – PΔV

ΔU = 800 × 10³ – 4.5 × 10⁵ (2 – 0.5) =1000(800 – 675)

⇒ dU = 125 kJ

Example:

As shown in figure when a system is taken from state a to state b, along the path    

a → c → b, 60 J of heat flow into the system, and 30 J of work is done:

(i) How much heat flows into the system along the path a → d → b if the work is 10 J.

(ii) When the system is returned from b to a along the curved path, the work done by the system is –20 J. Does the system absorb or liberate heat, and how much?

(iii) If, U_a = 0 and U_d = 22 J, find the heat absorbed in the process a → d and d → b.

Solution:

For the path acb ΔU = Q – W = 60 – 30 = 30 J or U_b – U_a = 30 J 

(i) Along the path adb, Q = ΔU + W = 30 + 10 = 40 J

(ii) Along the curved path ba, Q = (U_a – U_b) + W = (–30) + (–20) = –50 J, heat liberates from system 

(iii) Q_ad = U_d – U_a + W_ad 

but W_ad = W_adb – W_db = 10 – 0 = 10 Hence Q_ad = 22 – 0 + 10 = 32 J 

and Q_db = U_b – U_d + W_db = 30 – 22 + 0 = 8 J

5.0Limitations of the First Law of Thermodynamics

• No information about which part of heat is converted into mechanical work. 
• Does not give proper direction of heat flow.
• Does not tell about the feasibility of various thermodynamic processes.

Also Read: Second law of thermodynamics