Question

Which is correct for gases showing attraction behaviour?
A.${{\left( \dfrac{dV}{dP} \right)}_{T}}<0$
B.${{\left( \dfrac{d}{dP}PV \right)}_{T}}>0$
C.${{\left( \dfrac{d}{dP}PV \right)}_{T}}<0$
D.All of these

Answer 1

We know that the ideal gases are those gases which obey kinetic molecular theory of gases. In the ideal behaviour of gases, the volume occupied by gas molecules and forces of attraction between molecules is not considered.

Complete answer:
Real gases deviate from the behaviour of ideal gases since the real gas molecules do not satisfy mainly these two conditions of ideal gases. The theory was introduced to explain the behaviour of gases. The gas molecules were assumed as spheres in this theory and there were mainly two assumptions related to the kinetic theory of gases. When compared to the total volume occupied by the gas, the volume occupied by the gas molecules is negligibly small.
It is considered that there is no force of attraction between gas molecules and hence the collision between molecules are treated as elastic. Only at a higher temperature and lower pressure, the volume occupied by the gas molecules is negligible when compared to the total volume of gas. At lower temperatures and higher pressures, the molecules become compressed and the volume of gas molecules become significant Thus, the first assumption holds only at specific conditions. For one mole of gas $\Rightarrow Z=\dfrac{PV}{RT}$ ………(i)
Also, $Z=\dfrac{Observed\text{ }Molar\text{ }Volume}{Ideal\text{ }Molar\text{ }Volume},$ for attraction behaviour; $Z<1.$
Thus, by equating above equation and $Z<1$ we get; $\dfrac{PV}{RT}<1.$
At constant temperature (T) $PV < RT$ since $RT=constant.$
$P{{\left( \dfrac{dV}{dP} \right)}_{T}}+V{{\left( \dfrac{dV}{dP} \right)}_{T}}<0$ since $\dfrac{d}{dP}\left( constant \right)=0.$
$\Rightarrow P{{\left( \dfrac{dV}{dP} \right)}_{T}}+V<0$ here P cannot be negative, and volume also cannot be negative.
$\underset{\left[ +ve \right]}{\mathop{P}}\,{{\left( \dfrac{dV}{dP} \right)}_{T}}+\underset{\left[ +ve \right]}{\mathop{V}}\,<0$ since $PV < RT.$
Thus, ${{\left( \dfrac{d}{dP}PV \right)}_{T}}<0$ since $RT=constant$ and $\dfrac{d}{dP}RT=0$
Therefore, the correct answer is option C.

Note:
Remember that it should be noted that the completely ideal gas behaviour is a hypothetical concept. In ideal behaviour of gases, the gas molecules do not have any interaction and do not occupy space, as assumed in the kinetic theory of gases.