Question

1 \( \text{cm}^3 \) of water at its boiling point absorbs 540 cal of heat to become steam with a volume of 1671 \( \text{cm}^3 \). If the atmospheric pressure \( = 1.013 \times 10^5 \, \text{N/m}^2 \) and the mechanical equivalent of heat \( = 4.19 \, \text{J/cal} \), the energy spent in this process in overcoming intermolecular forces is

• A. 540 cal
• B. 40 cal
• C. 500 cal
• D. zero

Hint:

1 $cm$ of water at its boiling point absorbs 540 cal of heat to become steam with a volume of 1671 $cm$. If the atmospheric pressure $= 1.013x10 N/m$ and the mechanical equivalent of heat $= 4.19 J/cal$, the energy spent in this process in overcoming intermolecular forces is

Answer 1

The Correct Option is C

Step 1: First Law
$\Delta U = \Delta Q - \Delta W$.
Step 2: Work Done
$\Delta W = P\Delta V / J = \frac{1.013 \times 10^5 \times (1671 - 1) \times 10^{-6}}{4.2} \approx 40 \text{ cal}$.
Step 3: Internal Energy
$\Delta U = 540 - 40 = 500 \text{ cal}$.
Final Answer: (C)