Question

Calculate the work done against external atmospheric pressure when 1 g of water changes to
1672cm3
of steam. Take the atmospheric pressure as
1:013×105Nm−2

Answer 1

Answer on Question #41059, Physics, Molecular Physics | Thermodynamics

Calculate the work done against external atmospheric pressure when 1 g of water changes to 1672 cm³ of steam. Take the atmospheric pressure as 1:013×105Nm⁻²

Solution

Given:

$$P = 1.013 \cdot 10^5 \frac{\mathrm{N}}{\mathrm{m}^2}.$$ $$m = 1g = 10^{-3}kg.$$ $$V_2 = 1672 \mathrm{cm}^3 = 1.672 \cdot 10^{-2} \mathrm{m}^3.$$

1 g of water is evaporated under atmospheric pressure and temperature of 100 C. The density of steam at pressure $P = 1.013 \cdot 10^5 \frac{\mathrm{N}}{\mathrm{m}^2}$ and temperature of 100 C is $\rho_{\text{steam}} = 0.59 \frac{\mathrm{kg}}{\mathrm{m}^3}$. It takes volume

$$V_1 = \frac{m}{\rho_{\text{steam}}} = \frac{10^{-3}kg}{0.59 \frac{\mathrm{kg}}{\mathrm{m}^3}} = 1.69 \cdot 10^{-3} \mathrm{m}^3.$$

The work done against external atmospheric pressure is

$$W = P \Delta V = P (V_2 - V_1) = 1.013 \cdot 10^5 (1.672 \cdot 10^{-2} - 1.69 \cdot 10^{-3}) = 1.522 \mathrm{~kJ}.$$

Answer: 1.522 kJ.

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