
Gas $a/(K{{P}_{a}}d{{m}^{6}}mo{{l}^{-1}})$ $b/(d{{m}^{3}}mo{{l}^{-1}})$ A 642.32 0.05196 B 155.21 0.04136 C 431.91 0.05196 D 155.21 0.4382
a and b are van der waals constant. The correct statement about the gases are:
A. gas C will occupy less volume than gas A; gas B will be lesser compressible than gas D
B. gas C will occupy more volume than gas A; gas B will be lesser compressible than gas D
C. gas C will occupy more volume than gas A; gas B will be more compressible than gas D
D. gas C will occupy less volume than gas A; gas B will be more compressible than gas D
Gas | $a/(K{{P}_{a}}d{{m}^{6}}mo{{l}^{-1}})$ | $b/(d{{m}^{3}}mo{{l}^{-1}})$ |
A | 642.32 | 0.05196 |
B | 155.21 | 0.04136 |
C | 431.91 | 0.05196 |
D | 155.21 | 0.4382 |
Answer
516.6k+ views
Hint:. Van der Waals equation is basically a modified version of the Ideal Gas Law which states that gases consist of point masses which will undergo perfectly elastic collisions. But this law fails to explain the behavior of real gases that’s why Van der Waals equation was derived which helps us to define the physical state of a real gas.
Complete step by step answer:
Van der Waals equation gives us the relation between the pressure, volume, temperature, and amount of real gases. For a real gas containing ‘n’ moles, the equation can be written as:
$(P+\dfrac{a{{n}^{2}}}{{{V}^{2}}})(V-nb)=nRT$;
Where P = Pressure, V = Volume; T = Temperature; n = number of moles, b is the variable which is constant and relates the actual volume of the gas molecules and a is also a constant which is related with the strength of the attractive forces between gas molecules. Together the variable $\dfrac{a}{{{V}^{2}}}$ represents the correction factor for the pressure due to the attractive forces between gas molecules, while the variable (V-b) represents the correction factor for the volume due to the actual volume of the gas molecules themselves. Which defines that higher the, a value lesser will be the volume and pressure.
So, the correct answer is “Option C”.
Note: Van Der waal equation is able to predict the behavior of gases better than the ideal gas equation and it can also be applied on fluids. The cubic equation of van der waal equation gives three volumes that are useful for calculating the volume at and below critical temperatures. It is also able to calculate the critical conditions of liquefaction and derive an expression of the Principle of corresponding states.
Complete step by step answer:
Van der Waals equation gives us the relation between the pressure, volume, temperature, and amount of real gases. For a real gas containing ‘n’ moles, the equation can be written as:
$(P+\dfrac{a{{n}^{2}}}{{{V}^{2}}})(V-nb)=nRT$;
Where P = Pressure, V = Volume; T = Temperature; n = number of moles, b is the variable which is constant and relates the actual volume of the gas molecules and a is also a constant which is related with the strength of the attractive forces between gas molecules. Together the variable $\dfrac{a}{{{V}^{2}}}$ represents the correction factor for the pressure due to the attractive forces between gas molecules, while the variable (V-b) represents the correction factor for the volume due to the actual volume of the gas molecules themselves. Which defines that higher the, a value lesser will be the volume and pressure.
So, the correct answer is “Option C”.
Note: Van Der waal equation is able to predict the behavior of gases better than the ideal gas equation and it can also be applied on fluids. The cubic equation of van der waal equation gives three volumes that are useful for calculating the volume at and below critical temperatures. It is also able to calculate the critical conditions of liquefaction and derive an expression of the Principle of corresponding states.
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

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

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

Explain zero factorial class 11 maths CBSE
