JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Chemical Thermodynamics

Using the modern convention $\Delta U=q+w$, with $w$ the work done on the system, test each process. In an isochoric change no work is done, so $\Delta U=q$, correct. In a cyclic process the state returns to its start, so $\Delta U=0$ and $q=-w$, correct. In an isothermal change of an ideal gas the internal energy is unchanged, again giving $q=-w$, correct. In an adiabatic change $q=0$, so $\Delta U=w$; writing it as $\Delta U=-w$ reverses the sign and is the statement that does not correctly represent the first law.