Question

When water is boiled under a pressure of 2.00 atm, the heat of vaporization is 2.20×10^6 J/kg and the boiling point is 120 C. At this pressure, 1.00 kg of water has a volume of 1.00×10^−3 m3, and 1.00 kg of steam has a volume of 0.824 m^3.(A)Compute the work done when 1.00kg of steam is formed at this temperature(B)Compute the increase in internal energy of the water.

Answer 1

`\Delta U= 2.03 * 10^6 J`

Step-by-step explanation:

due to Isobaric system, the pressure is kept constant -

Work = Pressue( V_i - V_f)

WHERE

V_i IS INITIAL VOLUME

V_f is final volume

Work = 2.0*10^5 Pa (0.824 - 1.00x10^{-3})

Work = 1.64x10^5 J

B)

increase or decrease in internal energy can be calculated by using following relation:

`(\Delta U) = Q - W`

by keeping temperature and pressure constant, Q can be calculated from the phase change.

Q = (mass)(Heat of Vaporiztion)

Q = (1 kg)(2.20 *10^6) = 2.20 * 10^6 J

`(\Delta U) = Q - W`

`\Delta U= 2.20x10^6 J - 1.64x10^5 J`

`\Delta U= 2.03 * 10^6 J`