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The standard enthalpy of vapourisation of water at 100 degree Celsius is 40.63 kJ mol-1.the standard internal energy of vapourisation of water at 100 degree Celsius is? (A) 37.53 kJ mol-1 (B) 3.1 kJ mol-1 (C) 43.73 kJ mol-1 (D) 41.46 kJ mol-1 i need a perfect step solution. can any one help for me?
Most Upvoted Answer
The standard enthalpy of vapourisation of water at 100 degree Celsius ...
∆H=∆U+∆ngRT∆n=1 bcz h2o(l)-h2o(g)by putting appropriate values u get the ans as 37.53kjmol-
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The standard enthalpy of vapourisation of water at 100 degree Celsius ...
Calculation of Standard Internal Energy of Vaporization:
The standard enthalpy of vaporization (ΔHvap) is given as 40.63 kJ mol-1. To calculate the standard internal energy of vaporization (ΔUvap), we can use the equation:
ΔUvap = ΔHvap - PΔV
Where P is the pressure and ΔV is the change in volume.
Step 1: Determine the change in volume:
At 100°C, water is in its gaseous state (vapor) and occupies a larger volume compared to its liquid state. Therefore, the change in volume (ΔV) is positive.
Step 2: Determine the pressure:
The pressure is not given in the question. However, the standard conditions usually refer to a pressure of 1 bar or 1 atm. Therefore, we can assume the pressure to be 1 atm.
Step 3: Calculate the standard internal energy of vaporization:
Given:
ΔHvap = 40.63 kJ mol-1
P = 1 atm
We need to convert the pressure from atm to kJ mol-1. The conversion factor is 1 atm = 101.3 J mol-1.
ΔUvap = ΔHvap - PΔV
ΔUvap = 40.63 kJ mol-1 - (1 atm * ΔV * 101.3 J mol-1)
Step 4: Conversion of units:
To perform the calculation, we need to ensure that all units are consistent. Since the enthalpy of vaporization is given in kJ mol-1 and the pressure is in atm, we need to convert ΔV from L to m3.
Step 5: Conversion of volume units:
1 L = 0.001 m3
Step 6: Calculate the standard internal energy of vaporization:
Substitute the given values into the equation:
ΔUvap = 40.63 kJ mol-1 - (1 atm * ΔV * 101.3 J mol-1)
= 40.63 kJ mol-1 - (1 atm * ΔV * 101.3 J mol-1)
= 40.63 kJ mol-1 - (0.1013 kJ mol-1 * ΔV)
Since the change in volume (ΔV) is positive, the term - (0.1013 kJ mol-1 * ΔV) will be negative.
Step 7: Final calculation:
To determine the specific value of ΔUvap, we need additional information about the change in volume (ΔV). Without this information, we cannot accurately calculate the standard internal energy of vaporization.
Conclusion:
Without the knowledge of the change in volume (ΔV), we cannot determine the exact value of the standard internal energy of vaporization (ΔUvap). Therefore, we cannot choose any of the given options (A) 37.53 kJ mol-1, (B) 3.1 kJ mol-1, (C) 43.73 kJ mol-1, or (D)
The standard enthalpy of vaporization (ΔHvap) is given as 40.63 kJ mol-1. To calculate the standard internal energy of vaporization (ΔUvap), we can use the equation:
ΔUvap = ΔHvap - PΔV
Where P is the pressure and ΔV is the change in volume.
Step 1: Determine the change in volume:
At 100°C, water is in its gaseous state (vapor) and occupies a larger volume compared to its liquid state. Therefore, the change in volume (ΔV) is positive.
Step 2: Determine the pressure:
The pressure is not given in the question. However, the standard conditions usually refer to a pressure of 1 bar or 1 atm. Therefore, we can assume the pressure to be 1 atm.
Step 3: Calculate the standard internal energy of vaporization:
Given:
ΔHvap = 40.63 kJ mol-1
P = 1 atm
We need to convert the pressure from atm to kJ mol-1. The conversion factor is 1 atm = 101.3 J mol-1.
ΔUvap = ΔHvap - PΔV
ΔUvap = 40.63 kJ mol-1 - (1 atm * ΔV * 101.3 J mol-1)
Step 4: Conversion of units:
To perform the calculation, we need to ensure that all units are consistent. Since the enthalpy of vaporization is given in kJ mol-1 and the pressure is in atm, we need to convert ΔV from L to m3.
Step 5: Conversion of volume units:
1 L = 0.001 m3
Step 6: Calculate the standard internal energy of vaporization:
Substitute the given values into the equation:
ΔUvap = 40.63 kJ mol-1 - (1 atm * ΔV * 101.3 J mol-1)
= 40.63 kJ mol-1 - (1 atm * ΔV * 101.3 J mol-1)
= 40.63 kJ mol-1 - (0.1013 kJ mol-1 * ΔV)
Since the change in volume (ΔV) is positive, the term - (0.1013 kJ mol-1 * ΔV) will be negative.
Step 7: Final calculation:
To determine the specific value of ΔUvap, we need additional information about the change in volume (ΔV). Without this information, we cannot accurately calculate the standard internal energy of vaporization.
Conclusion:
Without the knowledge of the change in volume (ΔV), we cannot determine the exact value of the standard internal energy of vaporization (ΔUvap). Therefore, we cannot choose any of the given options (A) 37.53 kJ mol-1, (B) 3.1 kJ mol-1, (C) 43.73 kJ mol-1, or (D)
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The standard enthalpy of vapourisation of water at 100 degree Celsius is 40.63 kJ mol-1.the standard internal energy of vapourisation of water at 100 degree Celsius is? (A) 37.53 kJ mol-1 (B) 3.1 kJ mol-1 (C) 43.73 kJ mol-1 (D) 41.46 kJ mol-1 i need a perfect step solution. can any one help for me?
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The standard enthalpy of vapourisation of water at 100 degree Celsius is 40.63 kJ mol-1.the standard internal energy of vapourisation of water at 100 degree Celsius is? (A) 37.53 kJ mol-1 (B) 3.1 kJ mol-1 (C) 43.73 kJ mol-1 (D) 41.46 kJ mol-1 i need a perfect step solution. can any one help for me? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The standard enthalpy of vapourisation of water at 100 degree Celsius is 40.63 kJ mol-1.the standard internal energy of vapourisation of water at 100 degree Celsius is? (A) 37.53 kJ mol-1 (B) 3.1 kJ mol-1 (C) 43.73 kJ mol-1 (D) 41.46 kJ mol-1 i need a perfect step solution. can any one help for me? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The standard enthalpy of vapourisation of water at 100 degree Celsius is 40.63 kJ mol-1.the standard internal energy of vapourisation of water at 100 degree Celsius is? (A) 37.53 kJ mol-1 (B) 3.1 kJ mol-1 (C) 43.73 kJ mol-1 (D) 41.46 kJ mol-1 i need a perfect step solution. can any one help for me?.
The standard enthalpy of vapourisation of water at 100 degree Celsius is 40.63 kJ mol-1.the standard internal energy of vapourisation of water at 100 degree Celsius is? (A) 37.53 kJ mol-1 (B) 3.1 kJ mol-1 (C) 43.73 kJ mol-1 (D) 41.46 kJ mol-1 i need a perfect step solution. can any one help for me? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The standard enthalpy of vapourisation of water at 100 degree Celsius is 40.63 kJ mol-1.the standard internal energy of vapourisation of water at 100 degree Celsius is? (A) 37.53 kJ mol-1 (B) 3.1 kJ mol-1 (C) 43.73 kJ mol-1 (D) 41.46 kJ mol-1 i need a perfect step solution. can any one help for me? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The standard enthalpy of vapourisation of water at 100 degree Celsius is 40.63 kJ mol-1.the standard internal energy of vapourisation of water at 100 degree Celsius is? (A) 37.53 kJ mol-1 (B) 3.1 kJ mol-1 (C) 43.73 kJ mol-1 (D) 41.46 kJ mol-1 i need a perfect step solution. can any one help for me?.
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