Problem 157
Question
If \(Z\) is a compressibility factor, van der Waals equation at low pressure can be written as: (a) \(\mathrm{Z}=1-\frac{\mathrm{Pb}}{\mathrm{RT}}\) (b) \(Z=1+\frac{P b}{R T}\) (c) \(\mathrm{Z}=1+\frac{\mathrm{RT}}{\mathrm{Pb}}\) (d) \(Z=1+\frac{a}{V R T}\)
Step-by-Step Solution
Verified Answer
The correct answer is (b): \(Z=1+\frac{P b}{R T}\).
1Step 1: Understand the Van der Waals Equation
The van der Waals equation is used to describe the behavior of real gases: \[ \left(P + \frac{a}{V_m^2}\right)(V_m - b) = RT \] where \(P\) is the pressure, \(V_m\) is the molar volume, \(R\) is the gas constant, \(T\) is the temperature, and \(a\) and \(b\) are van der Waals constants.
2Step 2: Simplify for Low Pressure
At low pressure, the equation simplifies because the volume is very large, meaning \(b\) is negligible compared to \(V_m\) and the term \( \frac{a}{V_m^2} \) becomes very small compared to \(P\). Hence, the equation reduces to: \[ PV_m = RT \] which leads the compressibility factor \(Z\) to be: \( Z = \frac{PV}{RT} \).
3Step 3: Derive the Compressibility Factor Z
For a real gas at low pressure, the van der Waals equation is: \[ Z = \frac{PV_m}{RT} = 1 + \frac{Pb}{RT} \] assuming the term \( \frac{a}{V_m^2} \) is negligible. This equation shows how the volume correction \(b\) slightly modifies the ideal gas behavior.
4Step 4: Match with Given Options
Our derived equation \( Z = 1 + \frac{Pb}{RT} \) matches with option (b). The volume correction \(b\) shows a positive deviation from ideal behavior, which confirms the correct answer is option (b).
Key Concepts
Compressibility FactorReal GasesIdeal Gas Behavior
Compressibility Factor
The compressibility factor, denoted by the letter \( Z \), is a useful dimensionless quantity in the study of gases. It helps us understand how a real gas deviates from the ideal gas behavior. For an ideal gas, \( Z \) is equal to 1.Real gases, however, often do not behave in this ideal manner due to intermolecular forces and finite molecular volume. Therefore, the compressibility factor is expressed as:
\[ Z = \frac{PV}{RT} \]
- **P** represents pressure.- **V** is molar volume.- **R** stands for the ideal gas constant.- **T** is temperature.When \( Z > 1 \), it indicates the effects of molecular volume are dominant, meaning the molecules occupy significant space. Meanwhile, if \( Z < 1 \), attractive forces between molecules are stronger, causing the gas to occupy less volume than predicted by the ideal gas law.
\[ Z = \frac{PV}{RT} \]
- **P** represents pressure.- **V** is molar volume.- **R** stands for the ideal gas constant.- **T** is temperature.When \( Z > 1 \), it indicates the effects of molecular volume are dominant, meaning the molecules occupy significant space. Meanwhile, if \( Z < 1 \), attractive forces between molecules are stronger, causing the gas to occupy less volume than predicted by the ideal gas law.
- So, understanding \( Z \) helps chemists and scientists refine gas predictions beyond the simplicity of the ideal model.
Real Gases
Unlike ideal gases, real gases have volume and experience intermolecular forces. These differences become noticeable under conditions of high pressure or low temperature.
The van der Waals equation is often used to model these real gas behaviors. This equation incorporates two essential corrections:
- **Volume Correction (b):** Acknowledges that gas particles themselves take up space.
- **Pressure Correction (a):** Accounts for the intermolecular forces that reduce the actual pressure exerted by the gas.
Ideal Gas Behavior
In an ideal gas, the molecules do not interact with each other except for elastic collisions, and they occupy no volume. The ideal gas law, expressed as \( PV = nRT \), provides a simple linear relationship between the pressure (P), volume (V), and temperature (T) of a gas, with 'n' being the number of moles.
This law is based on assumptions:
This law is based on assumptions:
- The gas consists of many small particles placed far apart relative to their size.
- There are no attractive or repulsive forces between the particles.
- The collisions between particles and with container walls are completely elastic.
Other exercises in this chapter
Problem 155
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