
The Van der waals equation of the law of corresponding states for 1 mole of gas is:
(A) $({{P}_{r}}+\dfrac{3}{{{V}_{r}}^{2}})(3{{V}_{r}}-1)=8{{T}_{r}}$
(B) $({{P}_{r}}-\dfrac{3}{{{V}_{r}}^{2}})(3{{V}_{r}}+1)=8{{T}_{r}}$
(C) $({{P}_{r}}+\dfrac{3}{{{V}_{r}}^{2}})(3{{V}_{r}}+1)=8\pi {{T}_{r}}$
(D) $({{P}_{r}}+\dfrac{3}{{{V}_{r}}^{2}})(3{{V}_{r}}+1)=8$
Answer
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Hint: The van der waals equation for real gases is considered to be a corrected form of the ideal gas equation. In the van der waals equations new terms are introduced and the constant a and b are used.
Complete step by step solution:
The van der waals equation for the real gas is written as:
$(P+\dfrac{a{{n}^{2}}}{{{V}^{2}}})(V-nb)=nRT$
We have to find the van der Waals equation of the law of corresponding states for 1 mole of gas. The value of n= 1 the equation will be:
$(P+\dfrac{a}{{{V}^{2}}})(V-b)=RT$
In this question there are some extra terms we can write ${{P}_{r}}=\dfrac{P}{{{P}_{c}}}\text{ and }{{\text{V}}_{r}}=\dfrac{{{V}_{m}}}{{{V}_{c}}}\text{ and }{{\text{T}}_{r}}=\dfrac{T}{{{T}_{c}}}$
Therefore the value of ${{P}_{c}}=\dfrac{a}{27{{b}^{2}}},{{V}_{c}}=3b\text{ and }{{\text{T}}_{c}}=\dfrac{8a}{27Rb}$
After putting the values we get the equation as $({{P}_{r}}+\dfrac{3}{{{V}_{r}}^{2}})(3{{V}_{r}}-1)=8{{T}_{r}}$
Hence the correct answer is option (A).
Additional information:
In the van der waals equation we can see that new terms are included other than pressure, temperature, number of moles, temperature and universal gas constant. If we talk about the term $\dfrac{a{{n}^{2}}}{{{V}^{2}}}$ this is added for the pressure correction or the molecular attraction in the ideal gas law. We know that the p is pressure V is volume T is the temperature and n is the number of moles. This the constant a is related to the strength of attraction between the molecules of the gas.
Note: In the van der waals equation if we consider the second term that is nb. This term relates to the correction for the volume of the molecule of the molecule of a specific gas. So we can say that the constant b is related to the effective size or the molecules of the gases.
Complete step by step solution:
The van der waals equation for the real gas is written as:
$(P+\dfrac{a{{n}^{2}}}{{{V}^{2}}})(V-nb)=nRT$
We have to find the van der Waals equation of the law of corresponding states for 1 mole of gas. The value of n= 1 the equation will be:
$(P+\dfrac{a}{{{V}^{2}}})(V-b)=RT$
In this question there are some extra terms we can write ${{P}_{r}}=\dfrac{P}{{{P}_{c}}}\text{ and }{{\text{V}}_{r}}=\dfrac{{{V}_{m}}}{{{V}_{c}}}\text{ and }{{\text{T}}_{r}}=\dfrac{T}{{{T}_{c}}}$
Therefore the value of ${{P}_{c}}=\dfrac{a}{27{{b}^{2}}},{{V}_{c}}=3b\text{ and }{{\text{T}}_{c}}=\dfrac{8a}{27Rb}$
After putting the values we get the equation as $({{P}_{r}}+\dfrac{3}{{{V}_{r}}^{2}})(3{{V}_{r}}-1)=8{{T}_{r}}$
Hence the correct answer is option (A).
Additional information:
In the van der waals equation we can see that new terms are included other than pressure, temperature, number of moles, temperature and universal gas constant. If we talk about the term $\dfrac{a{{n}^{2}}}{{{V}^{2}}}$ this is added for the pressure correction or the molecular attraction in the ideal gas law. We know that the p is pressure V is volume T is the temperature and n is the number of moles. This the constant a is related to the strength of attraction between the molecules of the gas.
Note: In the van der waals equation if we consider the second term that is nb. This term relates to the correction for the volume of the molecule of the molecule of a specific gas. So we can say that the constant b is related to the effective size or the molecules of the gases.
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