Correct Answer: Option (B)

The incorrect combination is option (B). In a reversible adiabatic process for a
perfect (ideal) gas, the change in internal energy depends only on temperature and must be expressed using
the constant-volume heat capacity C_v, not C_p.

The correct relation is:

ΔU = ∫_T1^T₂ C_v(T) dT

Therefore, the use of C_p(T) in option (B) makes it incorrect.

Explanation of All Options

Option (A) – Constant Pressure Heating (No Phase Change)

At constant pressure, the mechanical work done by the system is:

w = −∫_V1^V₂ P dV

If the pressure is constant (isobaric process), this simplifies to:

w = −P(V₂ − V₁)

This expression is correct and matches the relation given in option (A).

Hence, option (A) is correct.

Option (B) – Reversible Adiabatic Process in a Perfect Gas

In an adiabatic process, heat exchange is zero:

q = 0

From the first law of thermodynamics:

ΔU = q + w = w

For an ideal gas, internal energy depends only on temperature:

U = U(T)

Therefore, the change in internal energy must be calculated using:

ΔU = ∫_T1^T₂ C_v(T) dT

Since option (B) incorrectly uses C_p(T), which is related to enthalpy changes,
option (B) is incorrect.

Option (C) – Reversible Isothermal Process in a Perfect Gas

For a reversible isothermal process:

w_rev = −∫_V1^V₂ P dV

Using the ideal gas equation:

P = nRT / V

The work expression becomes:

w_rev = −nRT ln(V₂/V₁)

Since the integral form given in the option is correct,
option (C) is a valid combination.

Option (D) – Constant Volume Heating (No Phase Change)

At constant volume:

dV = 0  ⇒  w = −∫ P dV = 0

The first law reduces to:

ΔU = q_V

For constant volume heating:

ΔU = ∫_T1^T₂ C_v dT

This relation is correct, so option (D) is also correct.

Final Conclusion

✔ The incorrect combination is Option (B).