First slide

Heat ;work and internal energy

Question

When water is boiled under a pressure of 2 atm, the heat of vaporization is $2.20 \times 10^{6} {Jkg}^{- 1}$ and the boiling point is 120°C. At this pressure, 1 kg of water has a volume of 10^-3 m³ and 1 kg of steam has a volume of 0.824 m³. What is the work done when 1 kg of steam is formed at this temperature?

Moderate

Solution

$Work done = p (Δ V) (at constant pressure)$
$\begin{array}{l} ∴ W = 2 atm \times (0 .824 - 0 .001) m^{3} \\ \Rightarrow W = 2 \times 1 .013 \times 10^{5} {Nm}^{- 2} \times 0 .823 m^{3} \\ = 166 .74 kJ \end{array}$

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