Two moles of oxygen is heated at a constant pressure from `0^(@)C`. What must be the gas for the volume to be doubled ? The specific heat of oxygen under these condition is `0.218 cal g^(-1) K^(-1)` .

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Understand the relationship between volume and temperature at constant pressure
At constant pressure, the volume of an ideal gas is directly proportional to its absolute temperature. This relationship can be expressed as:
\[ V \propto T \]
Using the ideal gas law, we can write:
\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]

### Step 2: Define the initial conditions
Given:
- Initial temperature \( T_1 = 0^\circ C = 273 \, K \)
- We want to double the volume, so \( V_2 = 2V_1 \)

### Step 3: Set up the equation for the final temperature
Using the relationship from Step 1:
\[ \frac{V_1}{T_1} = \frac{2V_1}{T_2} \]
Cancelling \( V_1 \) from both sides gives:
\[ \frac{1}{T_1} = \frac{2}{T_2} \]
Rearranging this gives:
\[ T_2 = 2T_1 \]

### Step 4: Calculate the final temperature
Substituting \( T_1 = 273 \, K \):
\[ T_2 = 2 \times 273 \, K = 546 \, K \]

### Step 5: Calculate the rise in temperature
The rise in temperature \( \Delta T \) is given by:
\[ \Delta T = T_2 - T_1 = 546 \, K - 273 \, K = 273 \, K \]

### Step 6: Calculate the mass of oxygen
The molar mass of oxygen (O₂) is approximately \( 32 \, g/mol \). For 2 moles:
\[ \text{Mass} (m) = 2 \, \text{moles} \times 32 \, g/mol = 64 \, g \]

### Step 7: Calculate the heat required
The heat required \( Q \) can be calculated using the formula:
\[ Q = m \cdot C_p \cdot \Delta T \]
Where:
- \( m = 64 \, g \)
- \( C_p = 0.218 \, \text{cal/g/K} \)
- \( \Delta T = 273 \, K \)

Substituting the values:
\[ Q = 64 \, g \cdot 0.218 \, \text{cal/g/K} \cdot 273 \, K \]

### Step 8: Perform the calculation
Calculating \( Q \):
\[ Q = 64 \cdot 0.218 \cdot 273 \]
\[ Q \approx 3809.344 \, \text{cal} \]

### Final Answer
The heat required to double the volume of 2 moles of oxygen when heated from \( 0^\circ C \) at constant pressure is approximately \( 3809 \, \text{cal} \).

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