All Exams Test series for 1 year @ ₹349 only
Question

From the above Carnot cycle undergone by an ideal gas, identify the processes in which the change in internal energy is NON-ZERO.

 

The correct answer is
II and IV

The Carnot cycle consists of four processes: two isothermal and two adiabatic. Let's analyze each process in the cycle to determine where the internal energy change is non-zero.

  1. Process I (W to X): Isothermal Expansion
    • In an isothermal process, the temperature remains constant.
    • For an ideal gas, the change in internal energy (\(\Delta U\)) is zero because internal energy depends only on temperature.
  2. Process II (X to Y): Adiabatic Expansion
    • In an adiabatic process, no heat is exchanged (\(Q = 0\)).
    • The work done results in a change in internal energy.
    • Thus, the change in internal energy (\(\Delta U \neq 0\)).
  3. Process III (Y to Z): Isothermal Compression
    • Again, the temperature remains constant.
    • The change in internal energy (\(\Delta U = 0\)).
  4. Process IV (Z to W): Adiabatic Compression
    • No heat is exchanged.
    • The work done results in a change in internal energy.
    • Thus, the change in internal energy (\(\Delta U \neq 0\)).

Therefore, the processes in which the change in internal energy is non-zero are II and IV, corresponding to the adiabatic processes.

Was this answer helpful?

Important Questions from First law of thermodynamics

  1. An ideal gas undergoes a process where $50J$ of work is done on the gas, and $30J$ of heat is removed from the gas.
    Which of the following statements is true for the gas?
  2. When a gas is compressed suddenly then its temperature

  3. During throttling process:

  4. Which among the following is correct option for adiabatic reversible expansion of an ideal gas?
  5. 2 mol of a monoatomic ideal gas with initial volume of 5 L and pressure 10 bar undergoes an irreversible adiabatic expansion against a constant final pressure of 1 bar. The final volume (in L) is ________. 
    (Given: R = $8.314 \times 10^{-2}$ L bar $mol^{-1}$ $K^{-1}$) 
    (rounded off to one decimal place)

Need Expert Advice?

Start Your Preparation with Prepp Mobile App

Download the app from Google Play & App Store
Download the app from Google Play & App Store
Prepp Mobile App