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

(a) 540 cals.

(b) 40 cals.

(c) 500 cals.

(d) zero

Ans: (c)

Sol: According to First Law of Thermodynamics

dQ = dU + dW

dU = dQ-dW

= mL – P(dV)

$ \displaystyle 1\times 540 – \frac{1.013 \times 10^5 (1671 – 1)\times 10^{-6}}{4.2} $

= 540-40

= 500 cals