To an ideal triatomic gas 800 cal heat energy is given at constant pressure . If vibrational mode is neglected , then energy used by gas in work done against surroundings is

Options

A. 200 cal
B. 300 cal
C. 400 cal
D. 60 cal

Answer

To solve the problem, we need to find the work done by an ideal triatomic gas when 800 calories of heat energy is added at constant pressure, neglecting vibrational modes.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Heat energy (q) = 800 calories
- The gas is an ideal triatomic gas (which has 3 degrees of freedom for translational and rotational motion, neglecting vibrational modes).
- The process occurs at constant pressure.

2. **Use the First Law of Thermodynamics**:
The First Law of Thermodynamics states:
\[ q = \Delta U + W \]
Where:
- \( q \) = heat added to the system
- \( \Delta U \) = change in internal energy
- \( W \) = work done by the system

Rearranging gives:
\[ W = q - \Delta U \]

3. **Calculate Change in Internal Energy (\(\Delta U\))**:
For an ideal gas, the change in internal energy at constant pressure can be expressed as:
\[ \Delta U = nC_V\Delta T \]
For a triatomic gas, the molar heat capacity at constant volume \( C_V \) is:
\[ C_V = \frac{3}{2}R + R = \frac{5}{2}R \]
Thus, we can express \(\Delta U\) in terms of \( n \) and \( \Delta T \):
\[ \Delta U = n \left(\frac{5}{2}R\right) \Delta T \]

4. **Calculate Work Done (\(W\))**:
At constant pressure, the work done by the gas can also be expressed as:
\[ W = P\Delta V \]
From the ideal gas law, we know:
\[ P\Delta V = nR\Delta T \]
Therefore, substituting for \(W\):
\[ W = nR\Delta T \]

5. **Relate \(\Delta U\) and \(W\)**:
From the earlier equations, we can express:
\[ \Delta U = \frac{5}{2}W \]
Thus, substituting this into the First Law equation:
\[ W = q - \frac{5}{2}W \]
Rearranging gives:
\[ W + \frac{5}{2}W = q \]
\[ \frac{7}{2}W = q \]
\[ W = \frac{2}{7}q \]

6. **Substitute the Value of \(q\)**:
Now substituting \( q = 800 \) calories:
\[ W = \frac{2}{7} \times 800 = \frac{1600}{7} \approx 228.57 \text{ calories} \]

### Final Answer:
The work done by the gas against the surroundings is approximately **228.57 calories**.