32 g of `O_(2)` is contained in a cubical container of side 1 m and maintained at a temperature of `127^(@)C`. The isothermal bulk modulus of elasticity of the gas is (universal gas constant = R)

Options

A. 127 R
B. 400 R
C. 200 R
D. 560 R

Answer

To find the isothermal bulk modulus of elasticity of the gas, we can follow these steps:

### Step 1: Identify the given data
- Mass of \( O_2 \) = 32 g
- Temperature = \( 127^\circ C \)
- Side of the cubical container = 1 m

### Step 2: Convert mass of \( O_2 \) to moles
The molar mass of \( O_2 \) is approximately 32 g/mol. Therefore, the number of moles \( n \) can be calculated as:
\[ n = \frac{\text{mass}}{\text{molar mass}} = \frac{32 \, \text{g}}{32 \, \text{g/mol}} = 1 \, \text{mol} \]

### Step 3: Convert temperature to Kelvin
To convert the temperature from Celsius to Kelvin:
\[ T = 127 + 273 = 400 \, K \]

### Step 4: Calculate the volume of the container
Since the container is a cube with a side length of 1 m, the volume \( V \) is:
\[ V = \text{side}^3 = 1^3 = 1 \, \text{m}^3 \]

### Step 5: Use the ideal gas law to find pressure
Using the ideal gas equation \( PV = nRT \), we can rearrange it to find pressure \( P \):
\[ P = \frac{nRT}{V} \]
Substituting the values:
- \( n = 1 \, \text{mol} \)
- \( R = R \) (universal gas constant)
- \( T = 400 \, K \)
- \( V = 1 \, \text{m}^3 \)

Thus,
\[ P = \frac{1 \cdot R \cdot 400}{1} = 400R \]

### Step 6: Relate bulk modulus to pressure
The isothermal bulk modulus \( \beta \) is proportional to the pressure:
\[ \beta \propto P \]
From the previous step, we can write:
\[ \beta = P = 400R \]

### Final Result
The isothermal bulk modulus of elasticity of the gas is:
\[ \beta = 400R \]