Question

At very high pressures, the compressibility factor of one mole of a gas is given by :

• A. $\frac{pb}{RT}$
• B. $1+ \frac{pb}{RT}$
• C. $ 1- \frac{pb}{RT}$
• D. $ 1- \frac{b}{(VRT)}$

Answer 1

The Correct Option is A

The given problem involves the computation of the compressibility factor of a gas at very high pressures. The compressibility factor (Z) for gases is a measure of how much the behavior of a real gas deviates from that of an ideal gas. At high pressures, the behavior of real gases diverges from the ideal gas law, and corrections are made using the Van der Waals equation.

The compressibility factor Z is defined as:

Z = \frac{pV_m}{RT}

Where:

• p is the pressure
• V_m is the molar volume
• R is the gas constant
• T is the temperature

According to Van der Waals equation, the real gas equation is:

\left(p + \frac{a}{V_m^2}\right) (V_m - b) = RT

At very high pressure, the volume correction term becomes significant, which results in the equation being dominated by the term involving 'b', the volume excluded by a mole of particles. Simplifying in the case of very high pressure:

V_m \approx b

Thus, plugging V_m = b into the definition of the compressibility factor Z:

Z = \frac{pb}{RT}

This result shows that at very high pressures, the compressibility factor (Z) is approximately \frac{pb}{RT}. This corresponds to the given correct answer.

Conclusion: The correct answer is \frac{pb}{RT} because at very high pressures, the compressibility factor of one mole of a gas is governed by the decrease in molar volume approximation by Van der Waals equation.