One gram of water (1cm³) becomes 1671 cm³ of steam when boiled at a constant pressure of 1 atm(1.013×10⁵pa). The heat of vaporization at this pressure is Lv=2.256×10⁶J/kg. Calculate:

1: the work done by water when it vaporizes
2: it increases internal energy

1 answer

To calculate the work done by the water when it vaporizes and the increase in internal energy, we can use the first law of thermodynamics, which states:

\[ \Delta U = Q - W \]

Where:

  • \(\Delta U\) is the change in internal energy,
  • \(Q\) is the heat added to the system,
  • \(W\) is the work done by the system.

Given Data:

  1. Volume of water (1 g) = \(1 , \text{cm}^3 = 1 \times 10^{-6} , \text{m}^3\)
  2. Volume of steam = \(1671 , \text{cm}^3 = 1671 \times 10^{-6} , \text{m}^3\)
  3. Heat of vaporization \(L_v = 2.256 \times 10^6 , \text{J/kg}\)
  4. Mass of water = \(1 , \text{g} = 0.001 , \text{kg}\)
  5. Pressure \(P = 1 , \text{atm} = 1.013 \times 10^5 , \text{Pa}\)

Step 1: Calculate the Work Done (W)

When water vaporizes, the work done by the water can be calculated using the formula:

\[ W = P \Delta V \]

Where:

  • \(\Delta V = V_{\text{steam}} - V_{\text{water}} = (1671 \times 10^{-6}) - (1 \times 10^{-6})\)
  • \(P\) is the pressure in pascals.

Calculating \(\Delta V\):

\[ \Delta V = 1671 \times 10^{-6} , \text{m}^3 - 1 \times 10^{-6} , \text{m}^3 = 1670 \times 10^{-6} , \text{m}^3 = 0.001671 , \text{m}^3 \]

Now, substituting this into the work done equation:

\[ W = (1.013 \times 10^5 , \text{Pa}) \times (0.001671 , \text{m}^3) \]

Calculating \(W\):

\[ W = 1.013 \times 10^5 \times 0.001671 = 169.72 , \text{J} \]

Step 2: Calculate the Heat Added (Q)

The heat added to vaporize the water is given by:

\[ Q = m L_v \]

Where:

  • \(m = 0.001 , \text{kg}\)
  • \(L_v = 2.256 \times 10^6 , \text{J/kg}\)

Calculating \(Q\):

\[ Q = 0.001 , \text{kg} \times 2.256 \times 10^6 , \text{J/kg} = 2260.256 , \text{J} \]

Step 3: Calculate the Increase in Internal Energy (\(\Delta U\))

Using the first law of thermodynamics:

\[ \Delta U = Q - W \]

Substituting the calculated values:

\[ \Delta U = 2260.256 , \text{J} - 169.72 , \text{J} = 2090.536 , \text{J} \]

Summary of Results:

  1. The work done by the water when it vaporizes: \(W \approx 169.72 , \text{J}\)
  2. The increase in internal energy: \(\Delta U \approx 2090.54 , \text{J}\)