exp the vaan der Waals equastion in real gas
Answers
Explanation:
.[1] The ideal gas law states that volume (V) occupied by n moles of any gas has a pressure (P) at temperature (T) in kelvins given by the following relationship, where R is the gas constant:
PV = nRT,
To account for the volume that a real gas molecule takes up, the van der Waals equation replaces V in the ideal gas law with {\displaystyle (V_{m}-b)}{\displaystyle (V_{m}-b)}, where Vm is the molar volume of the gas and b is the volume that is occupied by one mole of the molecules. This leads to:[1]
{\displaystyle P(V_{m}-b)=RT}{\displaystyle P(V_{m}-b)=RT}
The second modification made to the ideal gas law accounts for the fact that gas molecules do in fact attract each other and that real gases are therefore more compressible than ideal gases. Van der Waals provided for intermolecular attraction by adding to the observed pressure P in the equation of state a term {\displaystyle a/V_{m}^{2}}{\displaystyle a/V_{m}^{2}}, where a is a constant whose value depends on the gas. The van der Waals equation is therefore written as:[1]
{\displaystyle \left(P+a{\frac {1}{V_{m}^{2}}}\right)(V_{m}-b)=RT}{\displaystyle \left(P+a{\frac {1}{V_{m}^{2}}}\right)(V_{m}-b)=RT}
and can also be written as the equation below
{\displaystyle \left(P+a{\frac {n^{2}}{V^{2}}}\right)(V-nb)=nRT}{\displaystyle \left(P+a{\frac {n^{2}}{V^{2}}}\right)(V-nb)=nRT}
where Vm is the molar volume of the gas, R is the universal gas constant, T is temperature, P is pressure, and V is volume. When the molar volume Vm is large, b becomes negligible in comparison with Vm, a/Vm2 becomes negligible with respect to P, and the van der Waals equation reduces to the ideal gas law, PVm=RT.[1]
Answer:
ideal gas equation was pv=nRT,from this upon modification we get (p+an2/v2)(v_nb)=nRT
Explanation:
REFFER KHAN ACADEMY