Answer
Verified
377.7k+ views
Hint: To solve this question, we need to study the characteristics of real gases. According to the van der Waal equation for real gases a and b are constants of attraction and volume respectively. At low pressure we can neglect the comparison to a very large volume and thus we can find the correct option for the given graph.
Complete answer: Let,
According to a real gas equation, the constants 'a' and 'b' are Van der Waals constants for attraction 'a' and volume 'b' are characteristic constants for a given gas.
The 'a' value for a given gas is a measure of intermolecular forces of attraction. As the intermolecular forces of attraction increases, more will be the value of a.
For a given gas van der Waals constant of attraction 'a' is always greater than van der Waals constant of volume 'b'.
The gas having a higher value of ‘a’ can be liquefied easily and therefore hydrogen and Helium is not easily liquefiable.
Based on the above observations we can say that for gas A (Z > 1), a=0 and its dependence on P is linear at all pressures. Also, at high pressure, the slope is positive for all real gases and thus option D is also correct.
Thus we can say that the incorrect statement is Option (B).
Note:
Another way of solving this is by using the following method. From the graph it is clear that the value of z decreases with the increase of pressure. We can explain this using the van der Waals' equation. At high pressure, when the pressure is large, volume will be small and one cannot ignore 'b' in comparison to V. This means that z is greater than one and it increases linearly with pressure.
Complete answer: Let,
According to a real gas equation, the constants 'a' and 'b' are Van der Waals constants for attraction 'a' and volume 'b' are characteristic constants for a given gas.
The 'a' value for a given gas is a measure of intermolecular forces of attraction. As the intermolecular forces of attraction increases, more will be the value of a.
For a given gas van der Waals constant of attraction 'a' is always greater than van der Waals constant of volume 'b'.
The gas having a higher value of ‘a’ can be liquefied easily and therefore hydrogen and Helium is not easily liquefiable.
Based on the above observations we can say that for gas A (Z > 1), a=0 and its dependence on P is linear at all pressures. Also, at high pressure, the slope is positive for all real gases and thus option D is also correct.
Thus we can say that the incorrect statement is Option (B).
Note:
Another way of solving this is by using the following method. From the graph it is clear that the value of z decreases with the increase of pressure. We can explain this using the van der Waals' equation. At high pressure, when the pressure is large, volume will be small and one cannot ignore 'b' in comparison to V. This means that z is greater than one and it increases linearly with pressure.
Recently Updated Pages
A wire of length L and radius r is clamped rigidly class 11 physics JEE_Main
For which of the following reactions H is equal to class 11 chemistry JEE_Main
For the redox reaction MnO4 + C2O42 + H + to Mn2 + class 11 chemistry JEE_Main
In the reaction 2FeCl3 + H2S to 2FeCl2 + 2HCl + S class 11 chemistry JEE_Main
One mole of a nonideal gas undergoes a change of state class 11 chemistry JEE_Main
A stone is projected with speed 20 ms at angle 37circ class 11 physics JEE_Main
Trending doubts
Which is the longest day and shortest night in the class 11 sst CBSE
Who was the Governor general of India at the time of class 11 social science CBSE
Why is steel more elastic than rubber class 11 physics CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Define the term system surroundings open system closed class 11 chemistry CBSE
In a democracy the final decisionmaking power rests class 11 social science CBSE