Question:

An ideal gas undergoes a cyclic process as shown in Figure.

\(\Delta U_{BC}=-5\ \mathrm{kJ\ mol^{-1}}\), \(q_{AB}=2\ \mathrm{kJ\ mol^{-1}}\), \(W_{AB}=-5\ \mathrm{kJ\ mol^{-1}}\), \(W_{CA}=3\ \mathrm{kJ\ mol^{-1}}\). Heat absorbed by the system during process \(CA\) is :

Show Hint

U is a state function, so the sum of the three internal energy changes over the cycle is zero.
Updated On: Nov 11, 2026
  • \(-5\ \mathrm{kJ\ mol^{-1}}\)
  • \(+5\ \mathrm{kJ\ mol^{-1}}\)
  • \(18\ \mathrm{kJ\ mol^{-1}}\)
  • \(-18\ \mathrm{kJ\ mol^{-1}}\)
Show Solution
collegedunia
Verified By Collegedunia

The Correct Option is B

Solution and Explanation

The work sign convention (work done on the system is positive) agrees with the data: \(AB\) is an expansion at constant P so \(W_{AB}\) is negative, and \(CA\) is a compression so \(W_{CA}\) is positive.
\(\Delta U_{AB}=q_{AB}+W_{AB}=2-5=-3\) kJ/mol.
For a cycle \(\Delta U_{AB}+\Delta U_{BC}+\Delta U_{CA}=0\): \(-3-5+\Delta U_{CA}=0\Rightarrow\Delta U_{CA}=+8\) kJ/mol.
\(q_{CA}=\Delta U_{CA}-W_{CA}=8-3=+5\) kJ/mol.
Answer: (2) \(+5\ \mathrm{kJ\ mol^{-1}}\)
Was this answer helpful?
0
0