1 g 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 cm$^3$ during this phase change, then the increase in internal energy in the process will be :
Solution
Work done = P$\Delta$V<br/><br/>
= 3 × 10<sup>5</sup> × 1600 × 10<sup>–6</sup><br/><br/>
= 480 J<br/><br/>
Only 10% of heat is used in work done.<br/><br/>
Hence $\Delta$Q = 4800 J
The rest goes in internal energy, which is 90% of
heat.<br/><br/>
Change in internal energy = 0.9 × 4800 = 4320 J
About this question
Subject: Physics · Chapter: Thermodynamics · Topic: Zeroth and First Law
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