Set up and solve the Practice exmaple please. I understand the Primary example a

Question

Set up and solve the Practice exmaple please. I understand the Primary example and do not need help with that. Show all work. I do not understand how to solve for the W when I do not have Initial Pressure or Volume

One gram of water (1 cm3) becomes 1671 cm3 of steam when boiled at a constant pressure of 1 atm (1.013 times 105 Pa). The heat of vaporization at this pressure is Lv = 2.256 times 106 J / kg = 2256 J / g. Compute the work done by the water when it vaporizes; its increases in internal energy. For a constant - pressure we may use Equation 15.16, W = p(V2 - V1), to compute the work done by the vaporizing water. The work done by the vaporizing water is w = p(V2 - V1) = (1.013 times 105 Pa)(1671 times 10 -6 m3 -1 times 10 -6 m3) = 169 J. We find the heat added to the water and then use the first law of thermodynamics. The heat added to the water is the heat of vaporization, Q = mLv: = (1 g)(2256 J / g) = 2256 J. From the first law of thermodynamics, Equation 15.18, the change in internal energy is Delta U = Q - W 2256J - 169 J = 2087 J.

Explanation / Answer

you can calculate work done either by W = P*(V2 - V1)

or by first law of thermodynamics here.

please specify what do you want to know exactly.

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.