A cubical box of side `1 m` contains an ideal gas ata pressure `100 N//m^(2)`. If `Sigmav_(x)^(2) = Sigmav_(y)^(2) = Sigmav_(z)^(2) = 10^(28) m^(2)//s^(2)`, where `v_(x), v_(y)` and `v_(z)` are the x, y and z components, respectively, of the gas molecule, then the mass of each gas molecule is

Options

A. `10^(-20) g`
B. `10^(-23) g`
C. `10^(-18) g`
D. `10^(-26) g`

Explanation

To find the mass of each gas molecule in the cubical box, we can use the relationship between pressure, volume, and the average kinetic energy of the gas molecules. Here’s a step-by-step solution:

### Step 1: Understand the relationship between pressure, volume, and mass
We know that the pressure \( P \) of an ideal gas can be expressed in terms of the mass \( m \) of the gas molecules, the volume \( V \), and the average kinetic energy of the gas molecules. The formula is given by:

\[ P \cdot V = \frac{1}{3} \cdot m \cdot \sum v^2 \]

where \( \sum v^2 \) is the sum of the squares of the velocity components of the gas molecules.

### Step 2: Identify the given values
From the problem, we have:
- Pressure \( P = 100 \, \text{N/m}^2 \)
- Volume \( V = 1 \, \text{m}^3 \)
- \( \sum v_x^2 = \sum v_y^2 = \sum v_z^2 = 10^{28} \, \text{m}^2/\text{s}^2 \)

### Step 3: Calculate the total \( \sum v^2 \)
Since the components are equal, we can write:

\[ \sum v^2 = \sum v_x^2 + \sum v_y^2 + \sum v_z^2 = 3 \cdot 10^{28} \, \text{m}^2/\text{s}^2 \]

### Step 4: Substitute values into the equation
Now, substituting the values into the pressure-volume relationship:

\[ P \cdot V = \frac{1}{3} \cdot m \cdot \sum v^2 \]

Substituting the known values:

\[ 100 \cdot 1 = \frac{1}{3} \cdot m \cdot (3 \cdot 10^{28}) \]

### Step 5: Simplify the equation
This simplifies to:

\[ 100 = m \cdot 10^{28} \]

### Step 6: Solve for mass \( m \)
Now, we can solve for \( m \):

\[ m = \frac{100}{10^{28}} = 10^{-26} \, \text{kg} \]

### Step 7: Convert to grams
Since \( 1 \, \text{kg} = 1000 \, \text{grams} \):

\[ m = 10^{-26} \, \text{kg} = 10^{-23} \, \text{grams} \]

### Conclusion
Thus, the mass of each gas molecule is \( 10^{-26} \, \text{grams} \).