Question

2.-When water boils at 2 atm, the heat of vaporization is 2.2 106 J/kg, and the boiling point is 120ºC. At that pressure, 1 kg of steam occupies a volume of 0.824 m3, while 1 kg of water occupies 10-3 m3.
a) Find the work done when 1 kg of steam is formed at that temperature.
b) Calculate the increase in internal energy.

Answer 1

a) Find the work done when 1 kg of steam is formed at that temperature.

• p = pressure = 2 atm =2 101325.2738 = 202651 Pa
• V₁ = volume of 1kg of water = 10⁻³ m³
• V₂ = volume of 1 kg of steam = 0.824 m³
• Lf = heat of vaporization of water = 2.2 10⁶ J/kg
• m = mass of water = 1 kg

$\bf{W=p(V_{2}-V_{1})=166781 \ J }$

$\boxed{\bf{W=166781 \ J}}$

b) Calculate the increase in internal energy.

$\Delta U=Q-W\\Q=mL_{f}=2,2\cdot10^{6} \ J\\\Delta U=2,033\cdot10^{6} \ J\\\boxed{\Delta U=2,033\cdot10^{6} \ J}$