Question:easy
The thermodynamic process in which no work is done by the gas or on the gas is
The thermodynamic process in which no work is done by the gas or on the gas is
Show Hint
Work is P times the volume change, so a constant-volume process does no work.
Updated On: Oct 1, 2026
- isothermal process
- adiabatic process
- isobaric process
- isochoric process
Show Solution
The Correct Option is D
Solution and Explanation
Step 1: Idea:
A gas does work only by expanding or being compressed.
Step 2: Apply:
In an isochoric process the container is rigid. There is no displacement of the boundary, so no work.
Step 3: Answer:
Option (D).
Final Answer:
Option (D). \[ \boxed{\text{(D) isochoric process}} \]
A gas does work only by expanding or being compressed.
Step 2: Apply:
In an isochoric process the container is rigid. There is no displacement of the boundary, so no work.
Step 3: Answer:
Option (D).
Final Answer:
Option (D). \[ \boxed{\text{(D) isochoric process}} \]
Was this answer helpful?
0
Top Questions on Thermodynamics
- The difference of temperature in a material can convert heat energy into electrical energy. To harvest the heat energy, the material should have:
- JEE Main - 2025
- Physics
- Thermodynamics
- Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): With the increase in the pressure of an ideal gas, the volume falls off more rapidly in an isothermal process in comparison to the adiabatic process.
Reason (R): In an isothermal process, \( PV = \text{constant} \), while in an adiabatic process \( PV^\gamma = \text{constant} \). Here, \( \gamma \) is the ratio of specific heats, \( P \) is the pressure and \( V \) is the volume of the ideal gas.
In the light of the above statements, choose the correct answer from the options given below:- JEE Main - 2025
- Physics
- Thermodynamics
- A poly-atomic molecule (C\(_3\)R, \(C_v = 4R\), where \(R\) is gas constant) goes from phase space point A (\(P_A = 10^4 \, \text{Pa}, V_A = 4 \times 10^{-3} \, \text{m}^3\)) to point B (\(P_B = 5 \times 10^4 \, \text{Pa}, V_B = 6 \times 10^{-7} \, \text{m}^3\)) to point C (\(P_C = 10^4 \, \text{Pa}, V_C = 8 \times 10^{-3} \, \text{m}^3\)). A to B is an adiabatic path and B to C is an isothermal path. The net heat absorbed per unit mole by the system is:
- JEE Main - 2025
- Physics
- Thermodynamics
- If \[ \text{C(diamond)} \rightarrow \text{C(graphite)} + X \, \text{kj mol}^{-1} \] \[ \text{C(diamond)} + \text{O}_2(g) \rightarrow \text{CO}(g) + Y \, \text{kj mol}^{-1} \] \[ \text{C(graphite)} + \text{O}_2(g) \rightarrow \text{CO}(g) + Z \, \text{kj mol}^{-1} \] At constant temperature. Then:
- JEE Main - 2025
- Chemistry
- Thermodynamics
- Want to practice more? Try solving extra ecology questions todayView All Questions