Question

When a system is taken from state i to state f along the path iaf, it is found that \( Q = 50 \, \text{cal} \) and \( W = 20 \, \text{cal} \). Along the path ibf, \( Q = 36 \, \text{cal} \). W along the path ibf is:

• A. 14 cal
• B. 6 cal
• C. 16 cal
• D. 66 cal

Hint:

For cyclic processes, the change in internal energy is the same along different paths, allowing us to relate heat and work along various paths.

Answer 1

The Correct Option is C

The first law of thermodynamics states that: \[ \Delta U = Q - W \] Since the internal energy change \( \Delta U \) is the same for both paths, we can equate the two expressions for \( \Delta U \) and solve for \( W \) along path \( ibf \). This gives \( W = 16 \, \text{cal} \).
Final Answer: \[ \boxed{16 \, \text{cal}} \]