1 g of water, of volume `1 cm^3` at `100^@C` , is converted into steam at same temperature under normal atmospheric pressure ` (= 1 xx 10^5 Pa)` . The volume of steam formed equals `1671 cm^3` . If the specific latent heat of vaporisation of water is 2256 J/g, the change in internal energy is

Options

A. 2256 J
B. 2423 J
C. 2089 J
D. 167J

Explanation

To find the change in internal energy when 1 g of water at 100°C is converted into steam at the same temperature, we can use the formula:

\[ \Delta U = Q - W \]

Where:
- \(\Delta U\) is the change in internal energy,
- \(Q\) is the heat added to the system (in this case, the heat required to convert water to steam),
- \(W\) is the work done by the system.

### Step 1: Calculate the heat added (Q)
The heat added to convert water to steam can be calculated using the specific latent heat of vaporization:

\[ Q = m \cdot L \]

Where:
- \(m = 1 \, \text{g} = 0.001 \, \text{kg}\) (convert grams to kilograms for SI units),
- \(L = 2256 \, \text{J/g} = 2256000 \, \text{J/kg}\) (convert latent heat to J/kg).

Now substituting the values:

\[ Q = 1 \, \text{g} \cdot 2256 \, \text{J/g} = 2256 \, \text{J} \]

### Step 2: Calculate the work done (W)
The work done by the system when the steam expands can be calculated using the formula:

\[ W = P \cdot (V_f - V_i) \]

Where:
- \(P = 1 \times 10^5 \, \text{Pa}\) (normal atmospheric pressure),
- \(V_f = 1671 \, \text{cm}^3 = 1671 \times 10^{-6} \, \text{m}^3\) (final volume of steam),
- \(V_i = 1 \, \text{cm}^3 = 1 \times 10^{-6} \, \text{m}^3\) (initial volume of water).

Now substituting the values:

\[ W = 1 \times 10^5 \, \text{Pa} \cdot (1671 \times 10^{-6} \, \text{m}^3 - 1 \times 10^{-6} \, \text{m}^3) \]

Calculating the difference in volume:

\[ V_f - V_i = 1671 \times 10^{-6} - 1 \times 10^{-6} = 1670 \times 10^{-6} \, \text{m}^3 \]

Now substituting back into the work equation:

\[ W = 1 \times 10^5 \cdot 1670 \times 10^{-6} = 167 \, \text{J} \]

### Step 3: Calculate the change in internal energy (ΔU)
Now we can find the change in internal energy using the values of \(Q\) and \(W\):

\[ \Delta U = Q - W \]

Substituting the values:

\[ \Delta U = 2256 \, \text{J} - 167 \, \text{J} = 2089 \, \text{J} \]

### Final Answer:
The change in internal energy is:

\[ \Delta U = 2089 \, \text{J} \]