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Click here:point_up_2:to get an answer to your question :writing_hand:pick only the incorrect statement
Click here👆to get an answer to your question ✍️ Pick only the incorrect statement-for gas A- a-0-the compressibility factor is linearly dependent on pressure-for gas C-aneq 0-bneq 0-it can be used to calculate a and b by giving lowest P value-for gas B-0-if b-0-the compressibility factor is lineraly dependent on pressure-slope all three gases high pressure is positive
Solution- -C-xA0-for gas C-a-x2260-0-b-x2260-0- it can be used to calculate a and b by giving lowest P value-According to the real gas equation-The constants -apos-a-apos- and -apos-b-apos- are Van der Waals constant for attraction and volume for a given gas-The -apos-a-apos- values for a given gas are measure of intermolecular forces of attraction- More are the intermolecular forces of attraction- more will be the value of a-xA0-For a given gas van der Waals constant of attraction -apos-a-apos- is always greater than van der Waals constant of volume -apos-b-apos-xA0-The gas having higher value of -apos-a-apos-xA0- can be liquefied easily and therefore H2 and He are not liquefied easily-According to this- for gas A-Z-gt-1-a-0 and its dependence on P is linear at all pressure and for gas B-Z-lt-1-b-0 and its dependence on P is linear at all pressure-Also- at high pressure- the slope is positive for all real gases
Solved The van der Waals equation of state can be used to
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The given graph represent the variations of Z (compressibility factor (Z)=dfrac {pV}{nRT}) versus P, three real gases A, B and C. Identify the only incorrect statement.For the gas B, b=0 and its
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The given graph represents the variation of Z (compressibility factor =) versus P, for three real gases A, B and C. Identify the only incorrect statement. [JEE 2006]a)For the gas A, a =
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