Unit mass of liquid of volume `V_(1)` completely turns into a gas of volume `V_(2)` at constant atmospheric pressure P and temperature T. The latent heat of vaporization is "L". Then the change in internal energy of the gas is

Options

A. L
B. `L+p_(0)(V_(2)-V_(1))`
C. `L-p_(0)(V_(2)-V_(1)`
D. zero

Answer

To solve the problem, we need to determine the change in internal energy of the gas when a unit mass of liquid turns into gas at constant atmospheric pressure and temperature. Let's break down the solution step by step.

### Step 1: Understand the relationship between heat, work, and internal energy
We start with the first law of thermodynamics, which states:
\[ \Delta Q = \Delta U + \Delta W \]
Where:
- \(\Delta Q\) is the heat added to the system,
- \(\Delta U\) is the change in internal energy,
- \(\Delta W\) is the work done by the system.

### Step 2: Calculate the heat added to the system
The heat added to the system when a unit mass of liquid vaporizes is given by:
\[ \Delta Q = mL \]
Since we have a unit mass (m = 1), this simplifies to:
\[ \Delta Q = L \]

### Step 3: Calculate the work done by the system
The work done by the system during the phase change can be calculated using the formula:
\[ \Delta W = P \Delta V \]
Where:
- \(P\) is the constant atmospheric pressure,
- \(\Delta V\) is the change in volume, which is \(V_2 - V_1\).

Thus, we can write:
\[ \Delta W = P(V_2 - V_1) \]

### Step 4: Substitute values into the first law equation
Now, we can substitute the expressions for \(\Delta Q\) and \(\Delta W\) back into the first law equation:
\[ \Delta U = \Delta Q - \Delta W \]
Substituting the values we found:
\[ \Delta U = L - P(V_2 - V_1) \]

### Step 5: Final expression for change in internal energy
This gives us the final expression for the change in internal energy of the gas:
\[ \Delta U = L - P(V_2 - V_1) \]

### Summary
The change in internal energy of the gas when a unit mass of liquid completely turns into gas is given by:
\[ \Delta U = L - P(V_2 - V_1) \]