${100}^{\circ }C$ তাপমাত্রার 1 gm পানি ও ${100}^{\circ }C$ তাপমাত্রার 1 gm জলীয় বাষ্পের 1 বায়ুমন্ডলীয় চাপে এন্ট্রপির পার্থক্য কত ? ( ${100}^{\circ }C$ তাপমাত্রায় জলীয় বাষ্পের সুপ্ততাপ =540 cal/gm)

The provided answer of -1.45 cal/K is correct due to the principles of thermodynamics and entropy changes when comparing water and vapor at the same temperature.

When energy is added to water to convert it into steam, it results in a significant increase in entropy because the molecules move from a more ordered state (liquid) to a less ordered state (gas). Hence, the entropy of water at 100°C and the entropy of steam at 100°C differ due to the latent heat involved in the phase change.

Here's a brief explanation of how the calculation can lead to the given answer:

    
• The latent heat of vaporization at 100°C is given as 540 cal/gm, which means for every gram of water that turns into vapor at this temperature, 540 calories of heat energy is absorbed.
    
• The change in entropy (ΔS) can be calculated using the formula:     ΔS = Q/T     where Q is the heat added (in this case, the latent heat of vaporization) and T is the temperature in Kelvin.
    
• Since the temperature is 373 K (which is equivalent to 100°C), you would calculate the change in entropy between the two states, leading to the difference represented as -1.45 cal/K.

Thus, the answer provided reflects the correct understanding of entropy changes associated with the phase change at a constant temperature.

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