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Question: For the non-zero values of force of attraction between gas molecules, gas equation will be: A. \(P...

For the non-zero values of force of attraction between gas molecules, gas equation will be:
A. PV=nRTn2aVPV=nRT-\dfrac{{{n}^{2}}a}{V}
B. PV=nRTnbPPV=nRT-nbP
C. PV=nRTPV=nRT
D. P=nRTVbP=\dfrac{nRT}{V-b}

Explanation

Solution

The ideal gas equation is only applicable for the ideal gasses, and the real gasses does not obey the ideal gas equation in all temperature and pressure, in other words, the real gasses shows ideal behaviour in certain conditions of temperature and pressure.

The Vander Waals equation for real gases takes into account the attractive forces which are present between the particles of gases and so we add some pressure and volume correction terms in the original ideal gas equation.

Complete step by step answer:
The Ideal gas equation is a type of equation which is applicable only for ideal gases hypothetically. It has a number of limitations, as most of the gases deviate from their ideal behaviour when exposed to different conditions of temperature and pressure. It can be expressed as,
PV=nRTPV=nRT
Where PP is the pressure in which the gas is exposed, VV is the volume of the gas, nn represents the number of moles of the gas under consideration, RR is the gas constant, and TT represents the temperature at which the ideal gas is exposed. As we know that the particles of gases experience Vander Waals forces of attraction, but the ideal gas law doesn’t take into account those attractions. Moreover, if we look at all the options, we could eliminate option C as it is incorrect for the same reason.
Now we will write the Vander Waals equation which is applicable for real gasses,
(P+an2V2)(Vnb)=nRT(P+\dfrac{a{{n}^{2}}}{{{V}^{2}}})(V-nb)=nRT
Where PP is the pressure in which the gas is exposed, VV is the volume of the gas, nn represents the number of moles of the gas under consideration, RR is the gas constant, and TT represents the temperature at which the ideal gas is exposed, aa represents the intermolecular attractive forces which are present between the molecules or particles of the gas, and bb represents the effective size of the constituent gas particles. Here the an2V2\dfrac{a{{n}^{2}}}{{{V}^{2}}} is called the pressure correction term and the nbnb is called the volume correction term.
The Vander Waals constant bb can be calculated by using the formula,
b=4×NA×43πr3b=4\times {{N}_{A}}\times \dfrac{4}{3}\pi {{r}^{3}}
Where NA{{N}_{A}} Avogadro’s constant, and the formula for volume of sphere is 43πr3\dfrac{4}{3}\pi {{r}^{3}} where rr the radius of the sphere is.
Now if we consider the question, it says that the particles of gases have non-zero value of attractive forces which means that the value of aa cannot be neglected in the Vander Waals equation which is expressed above. And if the gas has a higher value of volume, the volume correction term can be neglected, meaning, we will neglect the term bb because it is a much smaller term than the volume of the gas. So now, we would get,
PV+an2V=nRTPV+\dfrac{a{{n}^{2}}}{V}=nRT
Now if we remove the bracket on the left side of the equation, we get,
(P+an2V2)V=nRT(P+\dfrac{a{{n}^{2}}}{{{V}^{2}}})V=nRT
Now we will keep the pressure and volume on the left side and take rest of the terms on the right side of the equation, we get,
PV=nRTan2VPV=nRT-\dfrac{a{{n}^{2}}}{V}
Now if we consider the options given in the question

So, the correct answer is Option A .

Note: The Vander Waals equation for real gases considers the attractive forces present between the constituent particles of the gases, so pressure and volume correction terms that is an2V2\dfrac{a{{n}^{2}}}{{{V}^{2}}} and nbnb is added to the original ideal gas equation and the modified equation becomes (P+an2V2)(Vnb)=nRT(P+\dfrac{a{{n}^{2}}}{{{V}^{2}}})(V-nb)=nRT. In other words, this equation has non-zero value of attractive forces.