The latent heat of vaporisation of water is 2240 J/gm. If the work done in the process of expansion of 1 g of water is 168 J, then increase in internal energy is

Options

A. 2408 J
B. 2072 J
C. 2240 J
D. 1904 J

Explanation

To solve the problem, we will use the first law of thermodynamics, which states that:

\[ Q = \Delta U + W \]

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

### Step 1: Identify the given values
- Latent heat of vaporization of water, \( Q = 2240 \, \text{J/g} \)
- Work done during the expansion, \( W = 168 \, \text{J} \)

### Step 2: Substitute the values into the equation
We need to find \( \Delta U \). Rearranging the first law of thermodynamics gives us:

\[ \Delta U = Q - W \]

Substituting the values we have:

\[ \Delta U = 2240 \, \text{J} - 168 \, \text{J} \]

### Step 3: Perform the calculation
Now, we calculate \( \Delta U \):

\[ \Delta U = 2240 - 168 = 2072 \, \text{J} \]

### Step 4: Conclusion
The increase in internal energy, \( \Delta U \), is:

\[ \Delta U = 2072 \, \text{J} \]