On the P–V diagram, the gas performs 5 J of work during isothermal path a→b and 4 J during adiabatic path b→c. Find change in internal energy (J) along direct path c→a.
Physics · Mixed Chapter
JEE Main Practice

Question

On the P–V diagram, the gas performs 5 J of work during isothermal path a→b and 4 J during adiabatic path b→c. Find change in internal energy (J) along direct path c→a.

G1 Physics · Accepted Answer

Step 1: For an ideal gas isothermal a→b, ΔU_ab=0. Step 2: For adiabatic b→c, Q_bc=0 and W_bc=+4 J, so ΔU_bc=Q−W=−4 J. Step 3: Net internal-energy change a→c is −4 J, independent of path. Step 4: Thus reverse c→a has ΔU_ca=+4 J.