A `3` mole sample of a triatomic ideal gas at `300 K` is allowed
to expand under adiabatic reversible condition from `5L` to `40 L`. The value of `DeltaH` is

Options

A. `-12.46 kJ`
B. `-14.965 kJ`
C. `-24.62 kJ`
D. `-10.24 kJ`

Explanation

To solve the problem of finding the change in enthalpy (ΔH) for a 3 mole sample of a triatomic ideal gas expanding adiabatically from 5L to 40L at 300K, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Number of moles (n) = 3 moles
- Initial temperature (T1) = 300 K
- Initial volume (V1) = 5 L
- Final volume (V2) = 40 L
- For a triatomic ideal gas, the specific heat at constant pressure (Cp) = 4R, where R = 8.314 J/(mol·K).

2. **Determine the Value of γ (gamma):**
- For a triatomic ideal gas, γ (gamma) = Cp/Cv. For triatomic gases, γ is typically around 1.33.

3. **Use the Adiabatic Condition:**
- The relationship for an adiabatic process is given by:
\[ T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1} \]
- Rearranging gives us:
\[ T_2 = T_1 \left(\frac{V_1}{V_2}\right)^{\gamma - 1} \]

4. **Calculate T2:**
- Substitute the known values into the equation:
\[ T_2 = 300 \left(\frac{5}{40}\right)^{1.33 - 1} \]
- Simplifying:
\[ T_2 = 300 \left(\frac{1}{8}\right)^{0.33} \]
- Calculate \( \left(\frac{1}{8}\right)^{0.33} \) which is approximately 0.5:
\[ T_2 \approx 300 \times 0.5 = 150 \text{ K} \]

5. **Calculate ΔH:**
- The change in enthalpy (ΔH) for an ideal gas is given by:
\[ \Delta H = n C_p (T_2 - T_1) \]
- Substitute the values:
\[ \Delta H = 3 \times 4R \times (150 - 300) \]
- Calculate:
\[ \Delta H = 3 \times 4 \times 8.314 \times (-150) \]
- This simplifies to:
\[ \Delta H = 3 \times 4 \times 8.314 \times (-150) = -14965.2 \text{ J} \approx -14.965 \text{ kJ} \]

6. **Final Answer:**
- The value of ΔH is approximately **-14.965 kJ**.