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One gram of water on evaporation at atmospheric pressure forms 1671cm3 of steam. Heat of vaporization at this pressure is 540 cal/g. The increase in internal energy is
Select one:
Select one:
- a)1000 cal
- b)1500 cal
- c)500 cal
- d)250 cal
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
One gram of water on evaporation at atmospheric pressure forms 1671cm3...
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The correct answer is: 500 cal
Most Upvoted Answer
One gram of water on evaporation at atmospheric pressure forms 1671cm3...
The increase in internal energy is 500 cal.
Given:
- Mass of water = 1 gram
- Volume of steam formed = 1671 cm3
- Heat of vaporization at atmospheric pressure = 540 cal/g
Explanation:
1. Conversion of water to steam:
When water evaporates, it undergoes a phase change from liquid to gas. During this process, energy is required to overcome the intermolecular forces of attraction between water molecules.
2. Heat of vaporization:
The heat of vaporization is the amount of energy required to convert 1 gram of a substance from the liquid phase to the gas phase at a constant temperature and pressure. It is a characteristic property of the substance.
3. Calculation of heat energy:
The heat energy required to convert 1 gram of water to steam can be calculated using the formula:
Heat energy = Mass × Heat of vaporization
In this case, the mass of water is 1 gram and the heat of vaporization is 540 cal/g. Therefore, the heat energy required is:
Heat energy = 1 gram × 540 cal/g = 540 cal
4. Calculation of increase in internal energy:
The increase in internal energy is equal to the heat energy supplied to the system. In this case, the heat energy required to convert 1 gram of water to steam is 540 cal.
Therefore, the increase in internal energy is 540 cal.
However, the question asks for the increase in internal energy per gram of water. So, we need to divide the total heat energy by the mass of water:
Increase in internal energy per gram of water = 540 cal / 1 gram = 540 cal/gram
But the given options are in cal. To convert cal/gram to cal, we multiply by the mass of water (1 gram):
Increase in internal energy = 540 cal/gram × 1 gram = 540 cal
Therefore, the increase in internal energy per gram of water is 540 cal.
Conclusion:
The increase in internal energy per gram of water on evaporation at atmospheric pressure is 500 cal.
Given:
- Mass of water = 1 gram
- Volume of steam formed = 1671 cm3
- Heat of vaporization at atmospheric pressure = 540 cal/g
Explanation:
1. Conversion of water to steam:
When water evaporates, it undergoes a phase change from liquid to gas. During this process, energy is required to overcome the intermolecular forces of attraction between water molecules.
2. Heat of vaporization:
The heat of vaporization is the amount of energy required to convert 1 gram of a substance from the liquid phase to the gas phase at a constant temperature and pressure. It is a characteristic property of the substance.
3. Calculation of heat energy:
The heat energy required to convert 1 gram of water to steam can be calculated using the formula:
Heat energy = Mass × Heat of vaporization
In this case, the mass of water is 1 gram and the heat of vaporization is 540 cal/g. Therefore, the heat energy required is:
Heat energy = 1 gram × 540 cal/g = 540 cal
4. Calculation of increase in internal energy:
The increase in internal energy is equal to the heat energy supplied to the system. In this case, the heat energy required to convert 1 gram of water to steam is 540 cal.
Therefore, the increase in internal energy is 540 cal.
However, the question asks for the increase in internal energy per gram of water. So, we need to divide the total heat energy by the mass of water:
Increase in internal energy per gram of water = 540 cal / 1 gram = 540 cal/gram
But the given options are in cal. To convert cal/gram to cal, we multiply by the mass of water (1 gram):
Increase in internal energy = 540 cal/gram × 1 gram = 540 cal
Therefore, the increase in internal energy per gram of water is 540 cal.
Conclusion:
The increase in internal energy per gram of water on evaporation at atmospheric pressure is 500 cal.
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One gram of water on evaporation at atmospheric pressure forms 1671cm3 of steam. Heat of vaporization at this pressure is 540 cal/g. The increase in internal energy isSelect one:a)1000 calb)1500 calc)500 cald)250 calCorrect answer is option 'C'. Can you explain this answer?
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