Question

One gram of a liquid is converted to vapour at \(3\times 10^5\) Pa pressure. If 10% of the heat supplied is used for increasing the volume by \(1600 \text{cm}^3\) during this phase change, then the increase in internal energy in the process will be

• A. \(4.32\times 10^8 \text{J}\)
• B. \(4800 \text{J}\)
• C. \(4320 \text{J}\)
• D. \(4.32\times 10^5 \text{J}\)

Hint:

Work done is P times the change in volume; it is 10 percent of the heat supplied.

Answer 1

The Correct Option is C

Step 1: Work done
\(W = P\Delta V = 3\times10^5\times1600\times10^{-6} = 480\ \text{J}\).

Step 2: Heat supplied
This work is \(10\%\) of the heat, so \(Q = 4800\ \text{J}\).

Step 3: First law
\(\Delta U = Q - W = 4800 - 480 = 4320\ \text{J}\). Option (C).

Step 4: Check
Option (B) is the total heat, not the change in internal energy.

Final Answer:
The increase in internal energy is 4320 J. \[ \boxed{\text{(C)}\ 4320\ \text{J}} \]