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: Split the heat
90 percent of the heat goes into internal energy and 10 percent into work.

Step 2: Compute
Work $= 480$ J is 10 percent, so heat $= 4800$ J and $\Delta U = 0.9\times4800 = 4320$ J. Option (C).

Final Answer:
4320 J. \[ \boxed{\text{(C)}\ 4320\ \text{J}} \]