Q160 / 200 · MHT CET - Medical 2026 (25 April Shift 2) · Physics
Identify the INCORRECT statement for isobaric process.
- A$\Delta p = 0$
- B$W \neq 0$
- C$Q \neq 0$
- Correct answerD$\Delta U = 0$
Solution
In an isobaric process, the pressure of the system remains constant, so $\Delta p = 0$.
The work done is given by $W = p\Delta V$. Since the volume changes, $W \neq 0$.
The heat exchanged is $Q = nC_p\Delta T$. Since the temperature changes, $Q \neq 0$.
The change in internal energy is $\Delta U = nC_v\Delta T$. Since the temperature changes, $\Delta U \neq 0$.
The statement $\Delta U = 0$ is true for an isothermal process of an ideal gas, not an isobaric process. Therefore, $\Delta U = 0$ is the incorrect statement.
Answer: $\Delta U = 0$
The work done is given by $W = p\Delta V$. Since the volume changes, $W \neq 0$.
The heat exchanged is $Q = nC_p\Delta T$. Since the temperature changes, $Q \neq 0$.
The change in internal energy is $\Delta U = nC_v\Delta T$. Since the temperature changes, $\Delta U \neq 0$.
The statement $\Delta U = 0$ is true for an isothermal process of an ideal gas, not an isobaric process. Therefore, $\Delta U = 0$ is the incorrect statement.
Answer: $\Delta U = 0$
Hence, the correct answer is option: 'D'.