1g of a liquid is converted to vapour at `3times 10^5` Pa presure. If 10% of the heat supplied is used for increasing the volume by `1600 cm^3` during this phase change increase in intenal energy in the process will be`

Options

A. 4800 J
B. 4320 J
C. `4.32times 10^(8)` J
D. 432000 J

Answer

To solve the problem step by step, we will follow the concepts of thermodynamics related to phase changes, work done, and internal energy.

### Step 1: Identify the given data
- Mass of the liquid, \( m = 1 \, \text{g} = 0.001 \, \text{kg} \)
- Pressure, \( P = 3 \times 10^5 \, \text{Pa} \)
- Volume change, \( \Delta V = 1600 \, \text{cm}^3 = 1600 \times 10^{-6} \, \text{m}^3 = 0.0016 \, \text{m}^3 \)
- Percentage of heat used for work, \( 10\% \)

### Step 2: Calculate the work done during the phase change
The work done \( W \) during the phase change can be calculated using the formula:
\[ W = P \Delta V \]
Substituting the values:
\[ W = (3 \times 10^5 \, \text{Pa}) \times (0.0016 \, \text{m}^3) = 480 \, \text{J} \]

### Step 3: Calculate the total heat supplied
Since 10% of the heat supplied is used for work done, we can express the total heat \( Q \) supplied as:
\[ 0.1Q = W \]
Thus,
\[ Q = \frac{W}{0.1} = \frac{480 \, \text{J}}{0.1} = 4800 \, \text{J} \]

### Step 4: Apply the first law of thermodynamics
According to the first law of thermodynamics:
\[ Q = \Delta U + W \]
Where \( \Delta U \) is the change in internal energy. Rearranging this gives:
\[ \Delta U = Q - W \]
Substituting the values we found:
\[ \Delta U = 4800 \, \text{J} - 480 \, \text{J} = 4320 \, \text{J} \]

### Final Answer
The increase in internal energy \( \Delta U \) during the process is:
\[ \Delta U = 4320 \, \text{J} \]

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