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 will follow these steps:

### Step 1: Convert Temperature to Kelvin
The given temperature is \(127^\circ C\). To convert this to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273 \]
Substituting the given value:
\[ T = 127 + 273 = 400 \, K \]

### Step 2: Calculate the Number of Moles of \(O_2\)
The molar mass of \(O_2\) (oxygen) is approximately \(32 \, g/mol\). To find the number of moles (\(n\)) in \(32 \, g\) of \(O_2\), we use the formula:
\[ n = \frac{\text{mass}}{\text{molar mass}} = \frac{32 \, g}{32 \, g/mol} = 1 \, mol \]

### Step 3: Determine the Volume of the Container
The volume of the cubical container is given as \(1 \, m^3\).

### Step 4: Use the Ideal Gas Law
The ideal gas law is given by:
\[ PV = nRT \]
Where:
- \(P\) = pressure
- \(V\) = volume
- \(n\) = number of moles
- \(R\) = universal gas constant (\(8.314 \, J/(mol \cdot K)\))
- \(T\) = temperature in Kelvin

### Step 5: Find the Pressure \(P\)
Rearranging the ideal gas law to find pressure:
\[ P = \frac{nRT}{V} \]
Substituting the known values:
\[ P = \frac{(1 \, mol)(8.314 \, J/(mol \cdot K))(400 \, K)}{1 \, m^3} = 3325.6 \, Pa \]

### Step 6: Calculate the Isothermal Bulk Modulus of Elasticity
The isothermal bulk modulus of elasticity (\(B\)) is given by:
\[ B = \frac{P}{V} \]
However, for ideal gases, the bulk modulus is also related to pressure:
\[ B = P \]
Thus, substituting the value of pressure we calculated:
\[ B = 3325.6 \, Pa \]

### Final Answer
The isothermal bulk modulus of elasticity of the gas is approximately:
\[ B \approx 3325.6 \, Pa \]