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What is change in internal energy when 2.0 mole of liquid water vaporises at 100 degree celsius? The heat of vapourisation of water at 100 degree celsius 40.66 kJ/mol.?
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What is change in internal energy when 2.0 mole of liquid water vapori...
We know that, for gaseous reactant and products, we have a relation between standard enthalpy of vaporization( ΔvapH0) and standard internal energy( ΔU0) as :
ΔvapH0 = ΔU0 + Δng RT
ΔvapH0=40.66KJ/mol = 2 * 40.66= 81.32 KJ for 2 moles of water.
,Δng = n2- n1 i.e. difference between no. of moles of gaseous reactant and gaseous product.
Here, for vaporization of water ,
2H2O (l) -> 2H2O (g)
So, change in no. of moles ,Δng = 2-0 =2
Thus, ΔvapH0 = ΔU0 +Δng RT
ΔU0 = ΔvapH0 - RT = 81.32 - ( 2 x 8.314 x 10-3 KJ/Kmol x 373K) = 81.32 - 6.2 KJ/mol = 75.12 KJ/mol
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What is change in internal energy when 2.0 mole of liquid water vapori...
Change in Internal Energy when Water Vaporizes

The change in internal energy when 2.0 moles of liquid water vaporize at 100 degrees Celsius can be calculated using the heat of vaporization of water at that temperature. The heat of vaporization of water at 100 degrees Celsius is given as 40.66 kJ/mol.

Explanation:

When a substance undergoes a phase change, such as vaporization, the internal energy of the substance changes. The internal energy is the sum of the kinetic and potential energies of the molecules in the substance. During a phase change, the internal energy does not change. Instead, it is used to break or form intermolecular forces between the molecules.

In the case of water vaporizing, the liquid water molecules gain enough energy to overcome the intermolecular forces and transition into the gaseous state. The heat of vaporization represents the amount of energy required to vaporize one mole of a substance at a given temperature.

To calculate the change in internal energy when 2.0 moles of liquid water vaporize at 100 degrees Celsius, we can use the following equation:

Change in Internal Energy = Number of Moles × Heat of Vaporization

Substituting the given values:

Change in Internal Energy = 2.0 mol × 40.66 kJ/mol

Calculating the result:

Change in Internal Energy = 81.32 kJ

Conclusion:

Therefore, the change in internal energy when 2.0 moles of liquid water vaporize at 100 degrees Celsius is 81.32 kJ. This value represents the energy required to overcome the intermolecular forces and convert the liquid water into water vapor at the given temperature.
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What is change in internal energy when 2.0 mole of liquid water vaporises at 100 degree celsius? The heat of vapourisation of water at 100 degree celsius 40.66 kJ/mol.?
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What is change in internal energy when 2.0 mole of liquid water vaporises at 100 degree celsius? The heat of vapourisation of water at 100 degree celsius 40.66 kJ/mol.? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about What is change in internal energy when 2.0 mole of liquid water vaporises at 100 degree celsius? The heat of vapourisation of water at 100 degree celsius 40.66 kJ/mol.? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for What is change in internal energy when 2.0 mole of liquid water vaporises at 100 degree celsius? The heat of vapourisation of water at 100 degree celsius 40.66 kJ/mol.?.
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