1 cm3 of water at its boiling point absorbs 540 cal of heat to become steam with a volume of 1671 cm3 . If the atmospheric pressure =1.013× 105 N/m2 and the mechanical equivalent of heat= 4.19J/cal, the energy spent in this process in overcoming intermolecular forces is

Question

1 cm3 of water at its boiling point absorbs 540 cal of heat to become steam with a volume of 1671 cm3 . If the atmospheric pressure =1.013× 105 N/m2 and the mechanical equivalent of heat= 4.19J/cal, the energy spent in this process in overcoming intermolecular forces is (A) 540 cal (B) 40 cal (C) 500 cal (D) zero

Tardigrade Admin · Accepted Answer

Change in volume =1671-1=1670 cm3 ∴ Work done (W)=p⋅ dV =(1.013× 105× 1670)J =(1670× 1.013× 105/4.2)cal =39.7cal Heat given (Q)=540 cal . From first law of thermodynamics, Δ U=Q-W =540-39.7 =500.3 cal ≈ 500 cal .

Q. 1cm3 of water at its boiling point absorbs 540 cal of heat to become steam with a volume of 1671cm3 . If the atmospheric pressure =1.013×105N/m2 and the mechanical equivalent of heat= 4.19J/cal, the energy spent in this process in overcoming intermolecular forces is

540 cal

40 cal

500 cal

zero

Change in volume =1671−1=1670cm3 ∴ Work done (W)=p⋅dV =(1.013×105×1670)J =4.21670×1.013×105​cal =39.7cal Heat given (Q)=540cal . From first law of thermodynamics, ΔU=Q−W =540−39.7 =500.3cal ≈500cal .