`H_(2)` gas is mixed with air at `25^(@)C` under a pressure of 1 atmosphere and exploded in a closed vessel. The heat of the reaction, `H_(2(g))+(1)/(2)O_(2(g))rarrH_(2)O_((v))` at constant volume, `DeltaU_("298 K")=-"240.60 kJ mol"^(-1)` and `C_(V)` values for `GH_(2)O` vapour and `N_(2)` in the temperature range `"298 K and 3200 K are 39.06 JK"^(-1)"mol"^(-1) and "26.40 JK"^(-1)"mol"^(-1)` respectively. The explosion temperature under adiabatic conditions is
(Given : `n_(N_(2))=2)`

Options

A. 2900 K
B. `2900^(@)C`
C. 2917 K
D. `3000^(@)C`

Explanation

To solve the problem of determining the explosion temperature under adiabatic conditions when hydrogen gas is mixed with air and exploded, we can follow these steps:

### Step 1: Understand the Reaction
The reaction occurring is:
\[ H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O(v) \]
The change in internal energy at 298 K for this reaction is given as:
\[ \Delta U_{298 K} = -240.60 \, \text{kJ/mol} \]

### Step 2: Set Up the Adiabatic Condition
Since the explosion occurs under adiabatic conditions, we know that:
\[ \Delta U = Q - W \]
For a closed vessel with no work done (isochoric process), we have:
\[ \Delta U = Q = 0 \]
Thus, the change in internal energy is equal to the heat released by the reaction:
\[ \Delta U = -\Delta U_{298 K} \]

### Step 3: Calculate the Heat Released
The heat released during the reaction can be expressed as:
\[ \Delta U = n \cdot C_v \cdot (T_f - T_i) \]
Where:
- \( n \) is the number of moles of gas involved,
- \( C_v \) is the total heat capacity at constant volume,
- \( T_f \) is the final temperature,
- \( T_i \) is the initial temperature (298 K).

### Step 4: Determine the Total Heat Capacity \( C_v \)
Given:
- \( C_v \) for \( H_2O \) (vapor) = 39.06 J/(K·mol)
- \( C_v \) for \( N_2 \) = 26.40 J/(K·mol)

Since we have 1 mole of \( H_2 \) and 0.5 moles of \( O_2 \) (which is negligible in terms of \( C_v \)), we consider:
- \( n_{N_2} = 2 \) moles (as given in the problem).

The total heat capacity \( C_v \) for the products can be calculated as:
\[ C_v = n_{H_2O} \cdot C_{v,H_2O} + n_{N_2} \cdot C_{v,N_2} \]
\[ C_v = 1 \cdot 39.06 + 2 \cdot 26.40 \]
\[ C_v = 39.06 + 52.80 = 91.86 \, \text{J/(K·mol)} \]

### Step 5: Relate Heat Capacity to Temperature Change
Now, substituting into the equation for heat released:
\[ 91.86 \cdot (T_f - 298) = 240600 \, \text{J/mol} \]
\[ T_f - 298 = \frac{240600}{91.86} \]
\[ T_f - 298 = 2615.59 \]
\[ T_f = 2615.59 + 298 \]
\[ T_f \approx 2913.59 \, \text{K} \]

### Final Step: Conclusion
Thus, the explosion temperature under adiabatic conditions is approximately:
\[ T_f \approx 2917 \, \text{K} \]