Second Law Of Thermodynamics MCQ Level - 1 (Part - 1) - Physics MCQ

By ideal gas equation,
pV = nRT     ...(i)
and     ...(ii)
Multiplying Equation (i) and (ii),
pVΔQ = nRΔS
⇒ 
∴ For isothermal expansion ΔV = 0

The correct answer is: (pVΔS)\(nR)

T₁ = (27 + 273)K = 300K
T₂ = (127 + 273)K = 400K
Efficiency,  

The correct answer is: 25%

 A Carnot cycle can be represented in a T-S diagram, with T along the y-axis and S (entropy) along the x-axis. The isotherms are parallel horizontal lines at Tc and Th. The isentropes are parallel vertical lines (by definition they are of constant entropy). Thus, a Carnot cycle is represented by a rectangle.

The entropy of an isolated system increases in an irreversible process

The second law of thermodynamics states that the entropy of an isolated system never decreases because isolated systems always evolve toward thermodynamic equilibrium, a state with maximum entropy.

Hence the entropy of an isolated system increases in an irreversible process.

Carnot cycle is a reversible process ideally and in reversible processes the entropy of universe remains constant instead of increasing as in an irreversible process.
 

 Answer :- d

Solution :- T1 = 127 + 273 = 400K

T2 = 227 + 273 = 500K

dT = T2 - T1

dT = 500 - 400 = 100

dS = dQ/dT

dS = 2000/100

dS = 20cal/K

Entropy is the measurement of Randomness of a System. Increase in thermal energy increases the randomness and hence Entropy. Now decreasing the temperature of a fluid results in decrease in it's thermal energy which decreases the randomness of the system which in turn decreases the entropy of the fluid. Now there is a finite randomness at normal temperature in the fluid and as we proceed from a higher temperature to a lower temperature the randomness decreases so does the entropy. After continuing this process when we reach 0K further no thermal energy can be extracted from the fluid and hence the system can be said to be completely ordered and as the system is ordered (i.e zero randomness) the entropy is assumed to be zero.

During the reversible process, the energy transfer as heat to the system from the surroundings is given by
Hence the area enclosed by a cycle on a T − S diagram represents the net work done by a system.
It also represents  

Efficiency of heat engine = T_source−T_sink  / T_source ( All temperatures in kelvin )
Hence , efficiency is 100% if T_sink = 0 K

• Entropy increases during an irreversible operation: Entropy is a measure of the disorder or randomness of a system. During irreversible operations, such as heat transfer from a hot body to a cold body, the disorder in the system increases, leading to an increase in entropy.


• Net change in entropy in a reversible cycle is zero: In a reversible cycle, the system undergoes a series of processes where the system is in equilibrium at every step. As a result, the entropy change for each step cancels out, leading to a net change in entropy of zero for the entire cycle.


• Change in entropy during an adiabatic operation is zero: An adiabatic process is one where there is no heat transfer between the system and its surroundings. Since entropy is related to heat transfer, in an adiabatic operation, there is no change in entropy.

All of the above properties are characteristics of entropy, highlighting its role in thermodynamics and energy transfer processes. Understanding these properties is essential for analyzing and predicting the behavior of systems in various physical and chemical processes.