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Question
One gram of water (1 cm3) becomes 1671 cm3 of steam at a pressure of 1 atm. The latent heat of vaporization at this pressure is 2256 J/g. Calculate the external work and the increase in internal energy.
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Solution
Given:
m = 1 g, Lvap = 2256 J/g,
p = 1.01 × 105 Pa, T = 100°C = 373 K,
Vsteam = 1671 cm3 , Vliq = 1 cm3
∴ ΔV = (Vsteam – Vliq) = (1671 – 1) = 1670 cm3 = 1670 × 10−6 m3
To find:
- External work done by the system (W)
- Increase in internal energy (ΔU)
Formulae:
- Q = mL
- W = pΔV
- ΔU = Q - W
Calculation:
From formula (i),
Q = 1 × 2256 = 2256 J
From formula (ii),
W = (1.01 × 105) × 1670 × 10−6
= 1687 × 10−1 J
≈ 169 J
From formula (iii),
ΔU = 2256 - 169 = 2087 J
- External work done by the system is 169 J.
- Increase in the internal energy is 2087 J.
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