Question

1 cm of water at 100°C and 1 atm pressure is
heated to convert it to 1521 cm3 vapour. Find
increase in internal energy if latent heat of
vaporization of water = 540 cal/g.
(1) 3.26 kJ
(2) 2.12 kJ
(3) 4.18 kJ
(4) 1.05 kJ
​

Answer 1

1$cm^3$ of water at 100°C and 1 atm pressure is heated to convert it to 1521$cm^3$ of vapour. The heat of vaporization of water =540 cal/g. Find the increase in internal energy.

Work done, W = Pressure * Change in Volume

= $1 atm * (1521 - 1) cm^3 =1.0125 * 10^5 Pa * 1520 * 10^{-6} m^3 = 153.9 J$

Heat required, Q = Mass * Latent Heat of vaporisation = $1 g * 540 cal/g = 540 cal = 540 * 4.18 J = 2257.2 J$

Increase in internal energy, ΔU = Q - W = 2103.3 J = 2.103 kJ

Hence option (2) is correct.

Answer 2

Thus the increase in internal energy is ΔU =  2.103 KJ

Option (2) is correct.

Explanation:

Given data:

• Volume of water = 1 cm
• Pressure = 1 atm
• Temperature = 100 °C
• New volume V2 = 1521 cm^3
• Latent heat of  vaporization of water = 540 cal/g.

Solution:

Work done = pressure x change in volume

W = P x ΔV

1 atm (1521 - 1 ) cm^3 = 1.0125 x 10^5 Pa x 1520 x 10^-6 m^3 = 153.9 J

Heat required "Q" = Mass x latent heat of vaporization

1 g x 540 cal / g = 540 cal

Q = 540 x 4.18 = 2257.2 J

Increase in internal energy ΔU = Q - W

ΔU = 2103.3 J = 2.103 KJ

Thus the increase in internal energy is ΔU =  2.103 KJ