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Question: Among the following, the correct statement is: A. At low pressure, real gases show ideal behavior....

Among the following, the correct statement is:
A. At low pressure, real gases show ideal behavior.
B. At very large volume, real gases show ideal behavior.
C. At Boyle’s temperature, real gases show ideal behavior.
D. At very low temperature, real gases show ideal behavior.

Explanation

Solution

We know that the Vander Waal equation is the combination of pressure, temperature, gas constant along with Vander Waal constants. Usually, the Vander Waal’s equation is affected by the volume and temperature.

Complete step by step answer
As we all know, the general form of Vander Waal’s equation can be written as:
[P+naV2](Vnb)=RT\left[ {P + \dfrac{{na}}{{{V^2}}}} \right]\left( {V - nb} \right) = RT
Where, the coefficient PP is the pressure of real gas, VV is the volume of real gas, RR is the gas constant, TT is the volume of real gas, nn is the number of moles and a,ba,b are the real gas constants.
The expression of Vander Waal’s equation for one moles of the real gas can be written as:
[P+aV2](Vb)=RT\left[ {P + \dfrac{a}{{{V^2}}}} \right]\left( {V - b} \right) = RT
So, when the volume of the real gas is very high the term aV2\dfrac{a}{{{V^2}}} will get neglected and hence, the final Vander Waal’s equation can be written as PV=RTPV = RT.
The general form of ideal gas equation can be written as PV=nRTPV = nRT.
Substitute the value of mole that is one in the above ideal gas general equation then we get the final equation which is similar to Vander Waal’s equation can be PV=RTPV = RT.
Thus, the obtained equation is the ideal gas equation for 1 mole of gas.

**Hence, the correct option for this given question is B that is at very large volume, real gases show ideal behavior.

Note:**
The “Ideal gas equation” can generally be utilized in the numerical part for calculating the unknown volumes of the gas along with the number of moles of the given gas. The general value of gas constant is always taken as 0.0821  L.atm/mol.K{\rm{0}}{\rm{.0821}}\;{\rm{L}}{\rm{.atm/mol}}{\rm{.K}} in terms of liters.