When water is boiled under a pressure of $$2$$ atm, the heat of vaporization is $$2.20$$×$$106Jkg$$−$$1$$ 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?

Question

When water is boiled under a pressure of $$2$$ atm, the heat of vaporization is $$2.20$$×$$106Jkg$$−$$1$$ 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?

Infinity Learn · Accepted Answer

Work done $$=p($$Δ$$V)$$ (at$$ constant $$pressure) ∴ W    $$=2atm$$×$$(0.824$$−$$0.001)m3$$⇒ W    $$=2$$×$$1.013$$×$$105Nm$$−$$2$$×$$0.823m3$$                $$=166.74kJ$$

Heat ;work and internal energy

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

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Heat ;work and internal energy

Remember concepts with our Masterclasses.

When water is boiled under a pressure of 2 atm, the heat of vaporization is 2.20×106Jkg−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?

Work done =p(ΔV) (at constant pressure) ∴ W =2atm×(0.824−0.001)m3⇒ W =2×1.013×105Nm−2×0.823m3 =166.74kJ

Ready to Test Your Skills?

free study material

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