Physics Question 33 – JEE-MAIN 2025

(A) Isobaric process: Pressure ($P$) is constant. The equation remains $\mathrm{\Delta }Q=\mathrm{\Delta }U+P\mathrm{\Delta }V$. This matches (IV). (B) Isochoric process: Volume ($V$) is constant, so $\mathrm{\Delta }V=0$. Thus, $\mathrm{\Delta }W=P\mathrm{\Delta }V=0$. The First Law becomes $\mathrm{\Delta }Q=\mathrm{\Delta }U$. This matches (II). (C) Adiabatic process: No heat exchange, so $\mathrm{\Delta }Q=0$. This matches (III). (D) Isothermal process: Temperature ($T$) is constant. For an ideal gas, internal energy ($U$) depends only on temperature, so $\mathrm{\Delta }U=0$. The First Law becomes $\mathrm{\Delta }Q=\mathrm{\Delta }W$. This matches (I).

Based on the analysis: (A) Isobaric $\to$ (IV) $\mathrm{\Delta }Q=\mathrm{\Delta }U+P\mathrm{\Delta }V$ (B) Isochoric $\to$ (II) $\mathrm{\Delta }Q=\mathrm{\Delta }U$ (C) Adiabatic $\to$ (III) $\mathrm{\Delta }Q=\text{zero}$ (D) Isothermal $\to$ (I) $\mathrm{\Delta }Q=\mathrm{\Delta }W$ This combination corresponds to option 4.