An ideal gas will have maximum density when

Options

A. `P = 0.5` atm , T = 600 K
B. P = 2 atm , T = 150 K
C. P = 1 atm , T = 300 K
D. P = 1.0 atm , T = 500 K

Explanation

To determine when an ideal gas will have maximum density, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Ideal Gas Law**:
The ideal gas law is given by the equation:
\[ PV = nRT \]
where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the gas constant, and \( T \) is temperature.

2. **Relate Density to the Ideal Gas Law**:
We know that density (\( \rho \)) can be expressed as:
\[ \rho = \frac{mass}{volume} = \frac{n \cdot M}{V} \]
where \( M \) is the molar mass of the gas. Substituting \( n = \frac{PV}{RT} \) into the density equation gives:
\[ \rho = \frac{PM}{RT} \]

3. **Identify the Relationship Between Density, Pressure, and Temperature**:
From the equation \( \rho = \frac{PM}{RT} \), we can see that:
\[ \rho \propto \frac{P}{T} \]
This means that density is directly proportional to pressure and inversely proportional to temperature.

4. **Maximize Density**:
To maximize density, we need to maximize \( P \) (pressure) and minimize \( T \) (temperature). Therefore, the conditions for maximum density occur when pressure is at its highest and temperature is at its lowest.

5. **Evaluate Given Options**:
Suppose we have four options with different pressures and temperatures. We need to identify which option has the highest pressure and the lowest temperature:
- Option A: 0.5 atm, 300 K
- Option B: 2 atm, 150 K
- Option C: 1 atm, 250 K
- Option D: 1 atm, 350 K

In this case:
- Option B has the highest pressure (2 atm) and the lowest temperature (150 K).

6. **Conclusion**:
Since option B has the highest \( P \) and the lowest \( T \), it will yield the maximum density. Therefore, the answer is:
\[ \text{Option B} \]