
‘a’ and ‘b’ are van der Waals constants for the gases. Chlorine is more easily liquefied than ethane because
(A) \$'a'\text{ for C}{{\text{l}}_{2}}>'a'for\text{ }{{\text{C}}_{2}}{{H}_{6}}\text{ but }\!\!'\!\!\text{ b }\!\!'\!\!\text{ for C}{{\text{l}}_{2}}<'b'\text{ for }{{\text{C}}_{2}}{{H}_{6}}$
(B) $'a'\text{ and }\!\!'\!\!\text{ b }\!\!'\!\!\text{ for C}{{\text{l}}_{2}}>'a'and\text{ }\!\!'\!\!\text{ b }\!\!'\!\!\text{ for }{{\text{C}}_{2}}{{H}_{6}}\text{ }$
(C) $'a'\text{ and }\!\!'\!\!\text{ b }\!\!'\!\!\text{ for C}{{\text{l}}_{2}}<'a'and\text{ }\!\!'\!\!\text{ b }\!\!'\!\!\text{ for }{{\text{C}}_{2}}{{H}_{6}}\text{ }$
(D) $'a'\text{ for C}{{\text{l}}_{2}}<'a'for\text{ }{{\text{C}}_{2}}{{H}_{6}}\text{ but }\!\!'\!\!\text{ b }\!\!'\!\!\text{ for C}{{\text{l}}_{2}}>'b'\text{ for }{{\text{C}}_{2}}{{H}_{6}}$
Answer
577.5k+ views
Hint: The van der Waals ‘a’ is a measure of the intermolecular forces while the van der Waals constant ‘b’ is the measure of molecular size. The value of van der Waals constant ‘a’ and ‘b’ are different for different types of gases.
Complete step by step solution:
The van der Waals equation is a better fit for the behavior of real gases than the ideal gas equation. The expression for the van der Waals equation for n mole of gases is mentioned below:
\[\left( P+\dfrac{a{{n}^{2}}}{{{V}^{2}}} \right)\left( V-nb \right)=nRT\]
Where $P$ = pressure of the gas
$V$ = volume of the gas
$T$ = temperature of the gas
$n$ = number of moles of the gas
‘$a$’ and ‘$b$’ are the van der waals constant.
The value of ‘a’ and ‘b’ for $C{{l}_{2}}\text{ and }{{\text{C}}_{2}}{{H}_{5}}$is given below:
For the gas greater the value of ‘a’ smaller will be the value of ‘b’. The value of van der Waals constant ‘a’ for chlorine is greater than the value of van der Waals constant ‘a’ for ethane so the value of van der Waals constant ‘b’ for chlorine will be less than the value of van der Waals constant ‘b’ for ethane.
Hence the correct answer is option (A).
Note: The van der Waals constant ‘a’ and ‘b’ are the characteristic properties of a particular gas. Larger the value of van der Waals constant ‘a’ stronger is the force of attraction between the particles. The value of van der Waals constant ‘a’ also relates to the boiling point of a gas.
Complete step by step solution:
The van der Waals equation is a better fit for the behavior of real gases than the ideal gas equation. The expression for the van der Waals equation for n mole of gases is mentioned below:
\[\left( P+\dfrac{a{{n}^{2}}}{{{V}^{2}}} \right)\left( V-nb \right)=nRT\]
Where $P$ = pressure of the gas
$V$ = volume of the gas
$T$ = temperature of the gas
$n$ = number of moles of the gas
‘$a$’ and ‘$b$’ are the van der waals constant.
The value of ‘a’ and ‘b’ for $C{{l}_{2}}\text{ and }{{\text{C}}_{2}}{{H}_{5}}$is given below:
| a | b | |
| $C{{l}_{2}}$ | 6.579${{L}^{2}}\text{ }bar\text{ }mo{{l}^{-2}}$ | 0.05622$L\text{ }mo{{l}^{-1}}$ |
| ${{\text{C}}_{2}}{{H}_{5}}$ | 5.562 ${{L}^{2}}\text{ }bar\text{ }mo{{l}^{-2}}$ | 0.06380 $L\text{ }mo{{l}^{-1}}$ |
For the gas greater the value of ‘a’ smaller will be the value of ‘b’. The value of van der Waals constant ‘a’ for chlorine is greater than the value of van der Waals constant ‘a’ for ethane so the value of van der Waals constant ‘b’ for chlorine will be less than the value of van der Waals constant ‘b’ for ethane.
Hence the correct answer is option (A).
Note: The van der Waals constant ‘a’ and ‘b’ are the characteristic properties of a particular gas. Larger the value of van der Waals constant ‘a’ stronger is the force of attraction between the particles. The value of van der Waals constant ‘a’ also relates to the boiling point of a gas.
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