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1 mole of ice at 0 degree Celsius 4.6 mm of HG pressure is converted to to water vapour at a constant temperature and pressure find the delta h and Delta e if the latent heat of fusion is 80 calorie per gram and latent heat of vaporization of liquid of water at zero degree celsius is fine and calorie per gram and volume of ice in comparison to that water is neglected?
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1 mole of ice at 0 degree Celsius 4.6 mm of HG pressure is converted t...
**Conversion of Ice to Water Vapor at Constant Temperature and Pressure**
To find the change in enthalpy (ΔH) and internal energy (ΔE) when 1 mole of ice at 0 degrees Celsius and 4.6 mm Hg pressure is converted to water vapor at the same temperature and pressure, we need to consider the different steps involved in the process.
**Step 1: Melting of Ice at 0°C**
Firstly, the ice needs to melt at 0 degrees Celsius. The latent heat of fusion (ΔHf) for ice is given as 80 calories per gram. Since we have 1 mole of ice, we need to convert the mass of ice to grams. The molar mass of water is approximately 18 g/mol.
Therefore, the mass of 1 mole of ice = 1 mole × 18 g/mol = 18 grams.
The energy required to melt the ice can be calculated using the formula: ΔH = ΔHf × mass
ΔH = 80 cal/g × 18 g = 1440 calories
**Step 2: Conversion of Water to Water Vapor at 0°C**
After the ice has melted, we have liquid water at 0 degrees Celsius. The next step is to convert this liquid water to water vapor at the same temperature and pressure.
The latent heat of vaporization (ΔHv) for liquid water at 0 degrees Celsius is given as 540 calories per gram. Again, we need to convert the mass of water to grams.
Using the same molar mass of water (18 g/mol) and assuming the density of water is approximately 1 g/mL, we can calculate the volume of water formed.
Volume of water = mass of water / density = 18 g / 1 g/mL = 18 mL
The energy required to vaporize the water can be calculated using the formula: ΔH = ΔHv × mass
ΔH = 540 cal/g × 18 g = 9720 calories
**Step 3: Calculation of ΔH and ΔE**
Since the process is occurring at constant temperature and pressure, the change in enthalpy (ΔH) is equal to the heat absorbed or released by the system. Therefore, the total change in enthalpy (ΔH) for the conversion of ice to water vapor is the sum of the enthalpy changes in Step 1 and Step 2.
ΔH = ΔH1 + ΔH2 = 1440 calories + 9720 calories = 11160 calories
Similarly, the change in internal energy (ΔE) can be calculated using the equation: ΔE = ΔH - PΔV, where PΔV is the work done by the system. Since the volume change is negligible, we can assume PΔV ≈ 0.
Therefore, ΔE = ΔH ≈ 11160 calories
Thus, the change in enthalpy and internal energy when 1 mole of ice at 0 degrees Celsius and 4.6 mm Hg pressure is converted to water vapor at the same temperature and pressure is approximately 11160 calories.
To find the change in enthalpy (ΔH) and internal energy (ΔE) when 1 mole of ice at 0 degrees Celsius and 4.6 mm Hg pressure is converted to water vapor at the same temperature and pressure, we need to consider the different steps involved in the process.
**Step 1: Melting of Ice at 0°C**
Firstly, the ice needs to melt at 0 degrees Celsius. The latent heat of fusion (ΔHf) for ice is given as 80 calories per gram. Since we have 1 mole of ice, we need to convert the mass of ice to grams. The molar mass of water is approximately 18 g/mol.
Therefore, the mass of 1 mole of ice = 1 mole × 18 g/mol = 18 grams.
The energy required to melt the ice can be calculated using the formula: ΔH = ΔHf × mass
ΔH = 80 cal/g × 18 g = 1440 calories
**Step 2: Conversion of Water to Water Vapor at 0°C**
After the ice has melted, we have liquid water at 0 degrees Celsius. The next step is to convert this liquid water to water vapor at the same temperature and pressure.
The latent heat of vaporization (ΔHv) for liquid water at 0 degrees Celsius is given as 540 calories per gram. Again, we need to convert the mass of water to grams.
Using the same molar mass of water (18 g/mol) and assuming the density of water is approximately 1 g/mL, we can calculate the volume of water formed.
Volume of water = mass of water / density = 18 g / 1 g/mL = 18 mL
The energy required to vaporize the water can be calculated using the formula: ΔH = ΔHv × mass
ΔH = 540 cal/g × 18 g = 9720 calories
**Step 3: Calculation of ΔH and ΔE**
Since the process is occurring at constant temperature and pressure, the change in enthalpy (ΔH) is equal to the heat absorbed or released by the system. Therefore, the total change in enthalpy (ΔH) for the conversion of ice to water vapor is the sum of the enthalpy changes in Step 1 and Step 2.
ΔH = ΔH1 + ΔH2 = 1440 calories + 9720 calories = 11160 calories
Similarly, the change in internal energy (ΔE) can be calculated using the equation: ΔE = ΔH - PΔV, where PΔV is the work done by the system. Since the volume change is negligible, we can assume PΔV ≈ 0.
Therefore, ΔE = ΔH ≈ 11160 calories
Thus, the change in enthalpy and internal energy when 1 mole of ice at 0 degrees Celsius and 4.6 mm Hg pressure is converted to water vapor at the same temperature and pressure is approximately 11160 calories.
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1 mole of ice at 0 degree Celsius 4.6 mm of HG pressure is converted t...
latent heat of vaporisation of liquid water at 0 degree celsius....?
iski koi value di hogi
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1 mole of ice at 0 degree Celsius 4.6 mm of HG pressure is converted to to water vapour at a constant temperature and pressure find the delta h and Delta e if the latent heat of fusion is 80 calorie per gram and latent heat of vaporization of liquid of water at zero degree celsius is fine and calorie per gram and volume of ice in comparison to that water is neglected?
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1 mole of ice at 0 degree Celsius 4.6 mm of HG pressure is converted to to water vapour at a constant temperature and pressure find the delta h and Delta e if the latent heat of fusion is 80 calorie per gram and latent heat of vaporization of liquid of water at zero degree celsius is fine and calorie per gram and volume of ice in comparison to that water is neglected? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 1 mole of ice at 0 degree Celsius 4.6 mm of HG pressure is converted to to water vapour at a constant temperature and pressure find the delta h and Delta e if the latent heat of fusion is 80 calorie per gram and latent heat of vaporization of liquid of water at zero degree celsius is fine and calorie per gram and volume of ice in comparison to that water is neglected? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1 mole of ice at 0 degree Celsius 4.6 mm of HG pressure is converted to to water vapour at a constant temperature and pressure find the delta h and Delta e if the latent heat of fusion is 80 calorie per gram and latent heat of vaporization of liquid of water at zero degree celsius is fine and calorie per gram and volume of ice in comparison to that water is neglected?.
1 mole of ice at 0 degree Celsius 4.6 mm of HG pressure is converted to to water vapour at a constant temperature and pressure find the delta h and Delta e if the latent heat of fusion is 80 calorie per gram and latent heat of vaporization of liquid of water at zero degree celsius is fine and calorie per gram and volume of ice in comparison to that water is neglected? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 1 mole of ice at 0 degree Celsius 4.6 mm of HG pressure is converted to to water vapour at a constant temperature and pressure find the delta h and Delta e if the latent heat of fusion is 80 calorie per gram and latent heat of vaporization of liquid of water at zero degree celsius is fine and calorie per gram and volume of ice in comparison to that water is neglected? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1 mole of ice at 0 degree Celsius 4.6 mm of HG pressure is converted to to water vapour at a constant temperature and pressure find the delta h and Delta e if the latent heat of fusion is 80 calorie per gram and latent heat of vaporization of liquid of water at zero degree celsius is fine and calorie per gram and volume of ice in comparison to that water is neglected?.
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