When unit mass of water boils to become steam at `100^(@)C`, it absorbs Q amount of heat. The densities of water and steam at `100^(@)C` are `rho_(1)` and `rho_(2)` respectively and the atmospheric pressure is `P_(0)`. The increase in internal energy of the water is

Options

A. a.`Q`
B. b.`Q +P_(0)((1)/(rho_(1)) - (1)/(rho_(2)))`
C. c.`Q + P_(0)((1)/(rho_(2)) - (1)/(rho_(1)))`
D. d.`Q - P_(0)((1)/(rho_(1)) + (1)/(rho_(2)))`

Explanation

To find the increase in internal energy of water when it boils to become steam at \(100^\circ C\), we can use the first law of thermodynamics, which states:

\[ \Delta U = \Delta Q - W \]

Where:
- \(\Delta U\) is the change in internal energy,
- \(\Delta Q\) is the heat absorbed,
- \(W\) is the work done by the system.

### Step 1: Identify the heat absorbed
Since it is given that the unit mass of water absorbs \(Q\) amount of heat to convert to steam, we can write:

\[ \Delta Q = Q \]

### Step 2: Calculate the work done
The work done \(W\) during the phase change at constant pressure can be expressed as:

\[ W = P \cdot (V_2 - V_1) \]

Where:
- \(V_2\) is the volume of steam,
- \(V_1\) is the volume of water.

### Step 3: Express volumes in terms of densities
Using the relationship between mass, volume, and density, we can express the volumes as:

\[ V_1 = \frac{m}{\rho_1} \quad \text{and} \quad V_2 = \frac{m}{\rho_2} \]

Since we are considering unit mass of water, \(m = 1\):

\[ V_1 = \frac{1}{\rho_1} \quad \text{and} \quad V_2 = \frac{1}{\rho_2} \]

### Step 4: Substitute volumes into the work done equation
Now substituting \(V_1\) and \(V_2\) into the work done equation:

\[ W = P_0 \left( \frac{1}{\rho_2} - \frac{1}{\rho_1} \right) \]

### Step 5: Substitute \(\Delta Q\) and \(W\) into the first law equation
Now we can substitute \(\Delta Q\) and \(W\) back into the first law of thermodynamics:

\[ \Delta U = Q - P_0 \left( \frac{1}{\rho_2} - \frac{1}{\rho_1} \right) \]

### Step 6: Rearranging the equation
This can be rearranged to:

\[ \Delta U = Q - P_0 \left( \frac{1}{\rho_2} - \frac{1}{\rho_1} \right) \]

### Final Expression
Thus, the increase in internal energy of the water when it boils to steam is:

\[ \Delta U = Q + P_0 \left( \frac{1}{\rho_1} - \frac{1}{\rho_2} \right) \]