Question:medium
\(1\) g of water of volume \(1\text{cm}^{3}\) at \(100^{\circ }\text{C}\) is converted into steam of same temperature under normal atmospheric pressure (\(10^{5}\) Pa). The volume of steam formed equals \(1671\text{cm}^{3}\). If the latent heat of vaporisation of water is \(2256\) J/g, the change in internal energy is
\(1\) g of water of volume \(1\text{cm}^{3}\) at \(100^{\circ }\text{C}\) is converted into steam of same temperature under normal atmospheric pressure (\(10^{5}\) Pa). The volume of steam formed equals \(1671\text{cm}^{3}\). If the latent heat of vaporisation of water is \(2256\) J/g, the change in internal energy is
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Heat supplied is the latent heat; subtract the work done in expanding against atmospheric pressure.
Updated On: Nov 12, 2026
- \(2423\) J
- \(2089\) J
- \(167\) J
- \(2256\) J
Show Solution
The Correct Option is B
Solution and Explanation
Step 1: Work Done by the System
The steam expands against constant pressure, so $W=P(V_{f}-V_{i})\approx 167$ J.
Step 2: Apply the First Law
$\Delta U=Q-W=2256-167$
Step 3: Result
$\Delta U=2089$ J, option (B).
Final Answer:
Subtracting the 167 J of expansion work leaves 2089 J. \[ \boxed{\text{(B) }2089\text{ J}} \]
The steam expands against constant pressure, so $W=P(V_{f}-V_{i})\approx 167$ J.
Step 2: Apply the First Law
$\Delta U=Q-W=2256-167$
Step 3: Result
$\Delta U=2089$ J, option (B).
Final Answer:
Subtracting the 167 J of expansion work leaves 2089 J. \[ \boxed{\text{(B) }2089\text{ J}} \]
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