3.3 First law of thermodynamics  (Page 4/6)

Vaporizing water

When 1.00 g of water at $100\text{°}\text{C}$ changes from the liquid to the gas phase at atmospheric pressure, its change in volume is $1.67\times {10}^{\mathrm{-3}}{\text{m}}^3\text{.}$ (a) How much heat must be added to vaporize the water? (b) How much work is done by the water against the atmosphere in its expansion? (c) What is the change in the internal energy of the water?

Strategy

We can first figure out how much heat is needed from the latent heat of vaporization of the water. From the volume change, we can calculate the work done from $W=p\text{Δ}V$ because the pressure is constant. Then, the first law of thermodynamics provides us with the change in the internal energy.

Solution

1. With $L_v$ representing the latent heat of vaporization, the heat required to vaporize the water is
   
   $Q=m{L}_v=(1.00\text{g})(2.26\times {10}^3\text{J/g})=2.26\times {10}^3\text{J}.$
2. Since the pressure on the system is constant at $1.00\text{atm}=1.01\times {10}^5{\text{N/m}}^2$ , the work done by the water as it is vaporized is
   
   $W=p\text{Δ}V=(1.01\times {10}^5{\text{N/m}}^2)(1.67\times {10}^{\mathrm{-3}}{\text{m}}^3)=169\text{J}\text{.}$
3. From the first law, the thermal energy of the water during its vaporization changes by
   
   $\text{Δ}{E}_{\text{int}}=Q-W=2.26\times {10}^3\text{J}-169\text{J}=2.09\times {10}^3\text{J}\text{.}$

Significance

We note that in part (c), we see a change in internal energy, yet there is no change in temperature. Ideal gases that are not undergoing phase changes have the internal energy proportional to temperature. Internal energy in general is the sum of all energy in the system.

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