The density or `O_2` gas at `127^o` C and 4.0 atm pressure is (R = 0.082 L atm `K^-1` `mol^-1`)

Options

A. 4.2 g/L
B. 3.9 g/L
C. 2.4 g/L
D. 1.5 g/L

Answer

To find the density of \( O_2 \) gas at \( 127^\circ C \) and \( 4.0 \, \text{atm} \), we can use the ideal gas law equation, which is given by:

\[ PV = nRT \]

Where:
- \( P \) = pressure (in atm)
- \( V \) = volume (in liters)
- \( n \) = number of moles
- \( R \) = ideal gas constant (\( 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \))
- \( T \) = temperature (in Kelvin)

### Step 1: Convert the temperature from Celsius to Kelvin
To convert Celsius to Kelvin, we use the formula:

\[ T(K) = T(°C) + 273.15 \]

For \( 127^\circ C \):

\[ T = 127 + 273.15 = 400.15 \, K \approx 400 \, K \]

### Step 2: Rearrange the ideal gas law to express density
The density (\( d \)) of a gas can be expressed as:

\[ d = \frac{m}{V} \]

Where \( m \) is the mass of the gas. The number of moles (\( n \)) can be expressed as:

\[ n = \frac{m}{M} \]

Where \( M \) is the molar mass of the gas. Substituting this into the ideal gas equation gives:

\[ PV = \frac{m}{M}RT \]

Rearranging for density:

\[ P = \frac{d \cdot V}{M}RT \]

Thus,

\[ d = \frac{PM}{RT} \]

### Step 3: Insert known values
For \( O_2 \):
- Molar mass \( M = 32 \, g/mol \)
- Pressure \( P = 4.0 \, atm \)
- Ideal gas constant \( R = 0.0821 \, L \, atm \, K^{-1} \, mol^{-1} \)
- Temperature \( T = 400 \, K \)

Substituting these values into the density formula:

\[ d = \frac{(4.0 \, atm)(32 \, g/mol)}{(0.0821 \, L \, atm \, K^{-1} \, mol^{-1})(400 \, K)} \]

### Step 4: Calculate the density
Calculating the numerator:

\[ 4.0 \times 32 = 128 \, g \cdot atm/mol \]

Calculating the denominator:

\[ 0.0821 \times 400 = 32.84 \, L \cdot atm/mol \]

Now, substituting these into the density formula:

\[ d = \frac{128}{32.84} \approx 3.9 \, g/L \]

### Final Answer
The density of \( O_2 \) gas at \( 127^\circ C \) and \( 4.0 \, atm \) is approximately \( 3.9 \, g/L \).

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