Answer
64.8k+ views
Hint: First check the Assertion with by observing their constants and conclude the result according to the Vanderwaals equation then go to reason and observe it also according to the Vanderwaals equation constants.
Complete Step by step Manner:
Let’s first talk about the Assertion we have here. In Assertion the Vanderwaals equation is
$\left[ P+\dfrac{a{{n}^{2}}}{{{V}^{2}}} \right]\left( V-nb \right)=RT$ in which $n$is number of moles of the gas. Constants$a$and$b$are called Vander waals constants and their values depend on the characteristic of a gas. And if we want to calculate the value of $'a'$ so it is a measure of magnitude of intermolecular attractive forces within the gas and is independent of temperature as pressure as we can see they are constants.
And also we cannot find any variance for $n$ so we can take it as 1($\therefore n=1$) so we will have our equation now $\left[ P+\dfrac{a}{{{V}^{2}}} \right]\left( V-b \right)=RT$
Here a and b are constants and they represent as:
$a:intermolecular\text{ }forces$
$b:co.volume$
Now when we see the equation we use to have which is $PV=nRT$ in which we use to assume that there is no force of attraction between them which means $a=0$. But we know that our gases can be liquefied which means they have intermolecular bonds between them. So, we can conclude that there are intermolecular forces between them. Now when we see our Assertion it says that Pressure correction $\left( a/{{V}^{2}} \right)$ is due to the force of attraction between molecules which now comes out to be true.
Now let’s talk about Reason which says that Volume of a gas molecule cannot be neglected due to force of attraction.
But now we see in the Assertion that both the constant have no relation between them on their own the constant $b:co.volume$ will exist independently so we can say that our Reason is false.
Hence, option C is the correct answer.
Note:
Here$'a'$ is the value of the intermolecular force of attraction for a given gas, so more the intermolecular forces of attraction than there will be more value of a. And for given gas the constant of attraction a is always greater than constant of volume b. And one interesting thing is that the gas which has a high value of a can be liquefied easily.
Complete Step by step Manner:
Let’s first talk about the Assertion we have here. In Assertion the Vanderwaals equation is
$\left[ P+\dfrac{a{{n}^{2}}}{{{V}^{2}}} \right]\left( V-nb \right)=RT$ in which $n$is number of moles of the gas. Constants$a$and$b$are called Vander waals constants and their values depend on the characteristic of a gas. And if we want to calculate the value of $'a'$ so it is a measure of magnitude of intermolecular attractive forces within the gas and is independent of temperature as pressure as we can see they are constants.
And also we cannot find any variance for $n$ so we can take it as 1($\therefore n=1$) so we will have our equation now $\left[ P+\dfrac{a}{{{V}^{2}}} \right]\left( V-b \right)=RT$
Here a and b are constants and they represent as:
$a:intermolecular\text{ }forces$
$b:co.volume$
Now when we see the equation we use to have which is $PV=nRT$ in which we use to assume that there is no force of attraction between them which means $a=0$. But we know that our gases can be liquefied which means they have intermolecular bonds between them. So, we can conclude that there are intermolecular forces between them. Now when we see our Assertion it says that Pressure correction $\left( a/{{V}^{2}} \right)$ is due to the force of attraction between molecules which now comes out to be true.
Now let’s talk about Reason which says that Volume of a gas molecule cannot be neglected due to force of attraction.
But now we see in the Assertion that both the constant have no relation between them on their own the constant $b:co.volume$ will exist independently so we can say that our Reason is false.
Hence, option C is the correct answer.
Note:
Here$'a'$ is the value of the intermolecular force of attraction for a given gas, so more the intermolecular forces of attraction than there will be more value of a. And for given gas the constant of attraction a is always greater than constant of volume b. And one interesting thing is that the gas which has a high value of a can be liquefied easily.
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