In which of the following cases a gas is more compressible than the ideal gas?

Options

A. `pbarVgtRT`
B. `pbarV=RT`
C. `pbarVltRT`
D. none

Answer

To determine in which case a gas is more compressible than an ideal gas, we need to analyze the compressibility factor (Z) of the gas. The compressibility factor is defined as:

\[ Z = \frac{PV}{nRT} \]

Where:
- \( P \) = pressure of the gas
- \( V \) = volume of the gas
- \( n \) = number of moles of the gas
- \( R \) = ideal gas constant
- \( T \) = temperature of the gas

### Step 1: Understand the Compressibility Factor
The compressibility factor (Z) indicates how much a real gas deviates from ideal gas behavior. For an ideal gas, \( Z = 1 \). If \( Z < 1 \), the gas is more compressible than an ideal gas, meaning it occupies less volume under the same conditions of temperature and pressure.

### Step 2: Identify Conditions for More Compressibility
To find cases where a gas is more compressible than an ideal gas, we need to look for conditions where the compressibility factor (Z) is less than 1. This typically occurs under certain conditions:
- High pressures
- Low temperatures
- Intermolecular forces are significant (e.g., in gases that are polar or have strong van der Waals forces)

### Step 3: Analyze the Given Cases
Now, we need to analyze the specific cases provided in the question (which are not listed here). For each case, we will determine if the compressibility factor (Z) is less than 1.

### Step 4: Conclusion
After analyzing all the cases, we can conclude which gas or condition results in a compressibility factor less than 1, indicating that the gas is more compressible than an ideal gas.