
Van der Waals constant b in a corrected equation for real gases represents:
(A)– measure the effective size of gas molecules.
(B)– magnitude of attractive forces among gas molecules.
(C)– free volume of the molecules
(D)– difference in pressure and volume of gas molecules
Answer
520.5k+ views
Hint: The constant a and b in the corrected real gas equation demonstrates the relation between pressure, temperature and volume and accounts for the deviation of real behaviour of gases from the ideal one. ‘a’ accounts for the pressure correction and ‘b’ accounts for volume correction.
Complete step by step answer:
First, let us know about the van der Waals equation for real gases. It is considered to be a corrected form given for an ideal gas.
Now, we will write the equation for an ideal gas, i.e. PV = nRT, and the van der Waals equation for real gases, i.e. (P + $\dfrac{an^{2}}{V^{2}}$ )(V-nb) = nRT.
In this equation, we can see the new terms are being introduced. If we talk about $\dfrac{an^{2}}{V^{2}}$ it is said to be the correction term for the pressure, or molecular attraction in the ideal gas law.
We know that P is pressure, V is the volume, T is the temperature, and n represents the no. of moles. Thus, a is the constant related to the strength of attraction between the molecules of a specific gas.
Now, the second correction term is nb. This term relates to the correction for the volume of molecules of a specific gas. So, we can say that the constant b is related to the effective size, or volume of the molecules.
In the last, we can conclude that the van der Waals constant b in a corrected equation for real gases presents a measure of the effective size of the gas molecules.
Hence, the correct option is (A).
Note: The modification was done as Van der Waal observed that ideal gas law fails to explain the behaviour of real gases. Ideal gas was based on the theory that gas molecules undergo perfectly elastic collisions. Also, the real gas equation is applicable for fluids as well.
The units of and b should be kept in mind-
Unit of a-$\text{atm li}{{\text{t}}^{2}}\text{mo}{{\text{l}}^{-2}}$and unit of b-$\text{lit mo}{{\text{l}}^{-1}}$
Complete step by step answer:
First, let us know about the van der Waals equation for real gases. It is considered to be a corrected form given for an ideal gas.
Now, we will write the equation for an ideal gas, i.e. PV = nRT, and the van der Waals equation for real gases, i.e. (P + $\dfrac{an^{2}}{V^{2}}$ )(V-nb) = nRT.
In this equation, we can see the new terms are being introduced. If we talk about $\dfrac{an^{2}}{V^{2}}$ it is said to be the correction term for the pressure, or molecular attraction in the ideal gas law.
We know that P is pressure, V is the volume, T is the temperature, and n represents the no. of moles. Thus, a is the constant related to the strength of attraction between the molecules of a specific gas.
Now, the second correction term is nb. This term relates to the correction for the volume of molecules of a specific gas. So, we can say that the constant b is related to the effective size, or volume of the molecules.
In the last, we can conclude that the van der Waals constant b in a corrected equation for real gases presents a measure of the effective size of the gas molecules.
Hence, the correct option is (A).
Note: The modification was done as Van der Waal observed that ideal gas law fails to explain the behaviour of real gases. Ideal gas was based on the theory that gas molecules undergo perfectly elastic collisions. Also, the real gas equation is applicable for fluids as well.
The units of and b should be kept in mind-
Unit of a-$\text{atm li}{{\text{t}}^{2}}\text{mo}{{\text{l}}^{-2}}$and unit of b-$\text{lit mo}{{\text{l}}^{-1}}$
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Trending doubts
1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

Explain zero factorial class 11 maths CBSE

What is the type of food and mode of feeding of the class 11 biology CBSE
