10 mole of an ideal gas is heated at constant pressure of one atmosphere from `27^(@)C` to `127^(@)C`. If `C_(v,m)=21.686+10^(-3)T(JK^(-1).mol^(-1))`, then `DeltaH` for the process is :

Options

A. 3000 J
B. 3350 J
C. 3700 J
D. `3181.4J, 2350J`

Explanation

To solve the problem, we need to calculate the change in enthalpy (ΔH) for the heating of an ideal gas at constant pressure. Here are the steps to find the solution:

### Step 1: Identify the given values
- Number of moles (n) = 10 moles
- Initial temperature (T1) = 27°C = 300 K (converted to Kelvin)
- Final temperature (T2) = 127°C = 400 K (converted to Kelvin)
- \( C_{v,m} = 21.686 + 10^{-3}T \) (in J/K·mol)

### Step 2: Calculate \( C_p \)
The relationship between \( C_p \) and \( C_v \) for an ideal gas is given by:
\[ C_p = C_v + R \]
Where \( R \) (the universal gas constant) is approximately \( 8.314 \, \text{J/K·mol} \).

Substituting the values:
\[ C_p = (21.686 + 10^{-3}T) + 8.314 \]
\[ C_p = 30 + 10^{-3}T \, \text{(in J/K·mol)} \]

### Step 3: Set up the formula for ΔH
The change in enthalpy (ΔH) at constant pressure can be calculated using the formula:
\[ \Delta H = n \int_{T_1}^{T_2} C_p \, dT \]

### Step 4: Substitute \( C_p \) into the integral
Substituting \( C_p \) into the equation:
\[ \Delta H = n \int_{T_1}^{T_2} (30 + 10^{-3}T) \, dT \]
\[ \Delta H = 10 \int_{300}^{400} (30 + 10^{-3}T) \, dT \]

### Step 5: Integrate the expression
Now, we can integrate:
\[ \Delta H = 10 \left[ 30T + \frac{10^{-3}}{2}T^2 \right]_{300}^{400} \]

Calculating the integral:
1. Evaluate at the upper limit (400 K):
\[ 30(400) + \frac{10^{-3}}{2}(400^2) = 12000 + \frac{10^{-3}}{2}(160000) = 12000 + 80 = 12080 \]

2. Evaluate at the lower limit (300 K):
\[ 30(300) + \frac{10^{-3}}{2}(300^2) = 9000 + \frac{10^{-3}}{2}(90000) = 9000 + 45 = 9045 \]

### Step 6: Calculate ΔH
Now, substituting these values back into the equation:
\[ \Delta H = 10 \left( 12080 - 9045 \right) \]
\[ \Delta H = 10 \times 3035 = 30350 \, \text{J} \]

### Final Answer
Thus, the change in enthalpy (ΔH) for the process is:
\[ \Delta H = 30350 \, \text{J} \]

---