
The heat of vaporization is high for?
A. He
B. Ne
C. A r
D. Xe
Answer
546.9k+ views
Hint: To solve this problem, firstly we need to understand what is heat of vaporization. It is defined as the amount of heat needed to turn $1$ gram of a liquid into a vapor, without a rise in the temperature of liquid.
Complete answer:
If the question is asking about the heat of vaporization, then we assume that all the gases are present in their liquid forms.
Now, we know vaporization in general and simple terms means the conversion of liquid to vapor.
And we know that whenever any liquid is converted to vapor, we determine it with the help of boiling point. So, we can infer that from the given noble gases the one with the highest boiling point will require more heat energy to get converted to vapor.
This simply means that the noble gas, which has the highest boiling point, would require more heat for vaporization and hence as a result it will have high heat of vaporization. So, we can conclude that heat of vaporization depends on boiling point.
We know that boiling point in turn depends on the van-der waal forces in case of noble gas. We should always remember the fact that van-der waal forces of that element will be greater which will have greater number of electrons and large size. From the trends of the periodic table, we know that as we move down the group an additional shell is added. As a result, the number of electrons and the size of element also increases.
So, we can conclude the trend of van-der waal forces for noble gases as;
He < Ne < Ar < Kr < Xe
From the above trend we can see that the van-der waal force of Xenon is greatest. As a result, the boiling point of xenon will be highest. Since, its boiling point is highest it will require more heat to undergo vaporization. Therefore, heat of vaporization is high for Xenon (Xe).
Hence, the correct answer is option ‘D’.
Note: It should be noted that heat of vaporization is generally measured in J/mol or kJ/mol (molar enthalpy of vaporization), although KJ/kg or J/g and older units like kcal/mol, calorie/g are sometimes still used among others. For faster calculation students can learn the following fact,
$\text{heat vaporization}\propto \text{Van-der wellforces}\propto\text{ Molar Mass}$
Complete answer:
If the question is asking about the heat of vaporization, then we assume that all the gases are present in their liquid forms.
Now, we know vaporization in general and simple terms means the conversion of liquid to vapor.
And we know that whenever any liquid is converted to vapor, we determine it with the help of boiling point. So, we can infer that from the given noble gases the one with the highest boiling point will require more heat energy to get converted to vapor.
This simply means that the noble gas, which has the highest boiling point, would require more heat for vaporization and hence as a result it will have high heat of vaporization. So, we can conclude that heat of vaporization depends on boiling point.
We know that boiling point in turn depends on the van-der waal forces in case of noble gas. We should always remember the fact that van-der waal forces of that element will be greater which will have greater number of electrons and large size. From the trends of the periodic table, we know that as we move down the group an additional shell is added. As a result, the number of electrons and the size of element also increases.
So, we can conclude the trend of van-der waal forces for noble gases as;
He < Ne < Ar < Kr < Xe
From the above trend we can see that the van-der waal force of Xenon is greatest. As a result, the boiling point of xenon will be highest. Since, its boiling point is highest it will require more heat to undergo vaporization. Therefore, heat of vaporization is high for Xenon (Xe).
Hence, the correct answer is option ‘D’.
Note: It should be noted that heat of vaporization is generally measured in J/mol or kJ/mol (molar enthalpy of vaporization), although KJ/kg or J/g and older units like kcal/mol, calorie/g are sometimes still used among others. For faster calculation students can learn the following fact,
$\text{heat vaporization}\propto \text{Van-der wellforces}\propto\text{ Molar Mass}$
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