A `3000 cm^(3)` tank contains oxygen at `20^(@)C` and the gauge pressure is `2.5 xx 10^(6) Pa`. Find the mass of the oxygen in the tank. Take 1 atm `= 10^(5) Pa`.

Explanation

To find the mass of oxygen in the tank, we can use the ideal gas equation, which is given by:

\[ PV = nRT \]

Where:
- \( P \) = pressure in pascals (Pa)
- \( V \) = volume in cubic meters (m³)
- \( n \) = number of moles of gas
- \( R \) = universal gas constant \( = 8.314 \, \text{J/(mol K)} \)
- \( T \) = temperature in Kelvin (K)

### Step 1: Convert the volume from cm³ to m³
The volume of the tank is given as \( 3000 \, \text{cm}^3 \). To convert this to cubic meters:

\[ V = 3000 \, \text{cm}^3 = 3000 \times 10^{-6} \, \text{m}^3 = 3.0 \times 10^{-3} \, \text{m}^3 \]

### Step 2: Convert the temperature from Celsius to Kelvin
The temperature is given as \( 20^\circ C \). To convert this to Kelvin:

\[ T = 20 + 273 = 293 \, \text{K} \]

### Step 3: Calculate the absolute pressure in the tank
The gauge pressure is given as \( 2.5 \times 10^6 \, \text{Pa} \). The absolute pressure \( P \) is the sum of the gauge pressure and the atmospheric pressure:

\[ P = P_0 + P_{\text{gauge}} = 10^5 \, \text{Pa} + 2.5 \times 10^6 \, \text{Pa} = 2.6 \times 10^6 \, \text{Pa} \]

### Step 4: Use the ideal gas equation to find the number of moles \( n \)
Rearranging the ideal gas equation to solve for \( n \):

\[ n = \frac{PV}{RT} \]

Substituting the known values:

\[ n = \frac{(2.6 \times 10^6 \, \text{Pa}) \times (3.0 \times 10^{-3} \, \text{m}^3)}{(8.314 \, \text{J/(mol K)}) \times (293 \, \text{K})} \]

Calculating \( n \):

\[ n = \frac{(2.6 \times 10^6) \times (3.0 \times 10^{-3})}{(8.314) \times (293)} \approx \frac{7800}{2437.282} \approx 3.20 \, \text{mol} \]

### Step 5: Calculate the mass of oxygen
The molar mass of oxygen \( O_2 \) is approximately \( 32 \, \text{g/mol} \). Therefore, the mass \( m \) can be calculated as:

\[ m = n \times \text{molar mass} = 3.20 \, \text{mol} \times 32 \, \text{g/mol} = 102.4 \, \text{g} \]

### Final Answer
The mass of the oxygen in the tank is approximately \( 102.4 \, \text{g} \).

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