The given question asks about the van der Waals equation at low pressure and its relation to the compressibility factor \(Z\). To solve this, let's discuss the concept of the compressibility factor and the van der Waals equation.
The van der Waals equation for real gases is expressed as:
\left(P + \frac{a}{V_m^2}\right)(V_m - b) = RTHere, \(P\) is the pressure, \(V_m\) is the molar volume, \(T\) is the temperature, \(R\) is the gas constant, and \(a\) and \(b\) are van der Waals constants specific to each gas. This equation accounts for the interactions between gas molecules and their volume.
The compressibility factor \(Z\) is defined as:
Z = \frac{PV_m}{RT}For an ideal gas, \(Z = 1\) always. Deviations from ideal behavior are expressed as \(Z \neq 1\), indicating interactions or volume exclusions.
At low pressures, the molar volume \(V_m\) is large, and we can simplify the van der Waals equation by neglecting terms involving \(b\), as the \(b\) term has less effect at large \(V\). The equation simplifies to:
PV_m = RT + \frac{a}{V_m}Substituting in the definition of \(Z\):
Z = \frac{PV_m}{RT} = 1 - \frac{a}{V_mRT}This matches the compressibility factor equation given in the options with respect to low-pressure scenarios. Therefore, the correct answer is:
Now, addressing why the other options are incorrect:
Thus, the correct option accurately reflects the adjustment due to intermolecular forces captured by the \(a\) term in the van der Waals equation under low pressure.