The compressibility factor for a real gas at high pressure is

Engineering

Chemistry

Vander Waal Gas and compressibility Factor

Liquefaction and Critical Temperature Pressure and Volume

Ideal Gas Equation

Question

The compressibility factor for a real gas at high pressure is

1 + RT/pb

1 + pb/RT

1 – pb/RT

At high pressure : $(P + \frac{a}{V^{2}}) \approx P$

$∴ (P + \frac{a}{V^{2}}) (V - b) = RT$

Reduce to P(V –b) = RT

or PV = RT + bP

$∴ Z = \frac{PV}{RT} = 1 + \frac{bP}{RT}$

The compressibility factor (Z) is a measure of how much a real gas deviates from ideal gas behavior. It is defined as:

$Z = \frac{P V}{n R T}$

For an ideal gas, Z = 1. For real gases, Z can be greater or less than 1 due to intermolecular forces and the finite size of gas molecules.

At high pressure, the volume is small, and the repulsive forces between molecules dominate. The van der Waals equation accounts for this with the 'b' term, which corrects for the volume occupied by the gas molecules themselves. The van der Waals equation is:

$(P + \frac{a n^{2}}{V^{2}}) (V - n b) = n R T$

At high pressure, the term $\frac{a n^{2}}{V^{2}}$ becomes negligible compared to the large pressure P. The equation simplifies to:

$P (V - n b) = n R T$

Solving for PV:

$P V = n R T + P n b$

Now, substitute this into the compressibility factor formula:

$Z = \frac{P V}{n R T} = \frac{n R T + P n b}{n R T} = \frac{n R T}{n R T} + \frac{P n b}{n R T} = 1 + \frac{P b}{R T}$

Therefore, the compressibility factor at high pressure is $1 + \frac{p b}{R T}$ .

Comparing with the given options, the correct answer is: $1 + \frac{p b}{R T}$

Detailed Solution

Related Topics and Formulae

Detailed Solution

Related Topics and Formulae

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