In a thermodynamic isobaric process:

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(C) An isobaric process is a thermodynamic process in which the pressure remains constant $(P = 	ext{constant})$.According to the first law of thermo

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change in internal energy is not zero

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(C) An isobaric process is a thermodynamic process in which the pressure remains constant $(P = 	ext{constant})$.According to the first law of thermodynamics, $\Delta Q = \Delta U + \Delta W$.In an isobaric process, work done is given by $\Delta W = P \Delta V$, which is generally not zero unless the volume change is zero.Heat supplied $\Delta Q = n C_p \Delta T$, which is not zero for a temperature change.The change in internal energy is given by $\Delta U = n C_v \Delta T$.Since $\Delta U$ depends on the change in temperature $(\Delta T)$, it is generally not zero in an isobaric process where the temperature changes.Therefore, the statement that the change in internal energy is not zero is correct.

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