Question

A gas described by van der Waals equation :
(A) behaves similar to an ideal gas in the limit of large molar volumes
(B) behaves similar to an ideal gas in the limit of large pressure
(C) is characterised by van der Waals coefficients that are dependent on the identity of gas but are independent of the temperature
(D) has the pressure that is lower than the pressure exerted by the same gas behaves ideally

Answer 1

Hint: The Van der Waals’s equation for state is given as -
$(P + \dfrac{a}{{{V^2}}})(V - b) = nRT$
Where $P$ = pressure of the gas
$V$ = volume of gas
$R$ = gas constant
$T$ = temperature
‘$a$’ and ‘$b$’ are critical constants
Thus, putting all the points in this equation, one can get the correct answer.

Complete step by step solution:
For such a question, we need to first understand what is the Van der Waals equation.
A scientist named J.D. van der Waals gave an equation of state. It is the modification of ideal gas law to explain the behaviour of real gases. This equation gives the relationship between pressure, volume, temperature and amount of real gases. The equation is -
$(P + \dfrac{a}{{{V^2}}})(V - b) = nRT$
Where $P$ = pressure of the gas
$V$ = volume of gas
$R$ = gas constant
$T$ = temperature
‘$a$’ and ‘$b$’ are critical constants
Now, let us see the options given to check which one is the correct answer.
The first option is that the gas behaves similar to an ideal gas in the limit of large molar volumes. If V is large then the term $\dfrac{a}{{{V^2}}}$is very small, it can be neglected. Further, as V is large, (V - b) will be equal to V. Thus, the equation will be equal to an ideal gas.
So, the option (A) is the correct answer.
The second option is that the gas behaves similar to an ideal gas in the limit of large pressure.
If the pressure is large, the term$\dfrac{a}{{{V^2}}}$is very small, it can be neglected. Further, (V - b) can not be neglected. Thus, the equation will not be equal to an ideal gas.
So, the option (B) is not the correct answer.
The third option is that the gas is characterised by van der Waals coefficients that are dependent on the identity of gas but are independent of the temperature. This is true because these coefficients depend on critical constants. These are independent of temperature.
So, the option (C) is the correct answer.
The fourth option is that the gas has the pressure that is lower than the pressure exerted by the same gas behaves ideally. This statement is true because we can see in van der Waals equation that a term $\dfrac{a}{{{V^2}}}$ is added to P. So, this option is even correct.

Thus, the correct options are (A), (C) and (D).

Note: The ideal gas equation is $PV =$ $nRT$
Where $P$ = pressure of the gas
$V$ = volume of gas
$R$ = gas constant
$T$ = temperature
All the gases that obey the ideal gas equation are called ideal gases. But in actual practice, no gas obeys the ideal gas equation and thus this equation was modified for the real gases.