For silver, `C_p(JK^(-1)mol^(-1))=23+0.01T` If the temperature (T) of 3 moles of silver is raised from 300 K to 1000 K at 1 atm pressure, the value of 'Delta H' will be close to

Options

A. 21 kJ
B. 13 kJ
C. 16 kJ
D. 62 kJ

Explanation

To find the change in enthalpy (ΔH) for the given process, we can use the formula:

\[ \Delta H = n \int_{T_1}^{T_2} C_p \, dT \]

Where:
- \( n \) = number of moles of silver = 3 moles
- \( C_p = 23 + 0.01T \) (given)
- \( T_1 = 300 \, K \)
- \( T_2 = 1000 \, K \)

### Step 1: Set up the integral

We need to calculate:

\[ \Delta H = 3 \int_{300}^{1000} (23 + 0.01T) \, dT \]

### Step 2: Break the integral into two parts

We can separate the integral into two parts:

\[ \Delta H = 3 \left( \int_{300}^{1000} 23 \, dT + \int_{300}^{1000} 0.01T \, dT \right) \]

### Step 3: Calculate the first integral

The first integral is:

\[ \int_{300}^{1000} 23 \, dT = 23[T]_{300}^{1000} = 23(1000 - 300) = 23 \times 700 = 16100 \]

### Step 4: Calculate the second integral

The second integral is:

\[ \int_{300}^{1000} 0.01T \, dT = 0.01 \left[\frac{T^2}{2}\right]_{300}^{1000} = 0.01 \left(\frac{1000^2}{2} - \frac{300^2}{2}\right) \]

Calculating the values:

\[ = 0.01 \left(\frac{1000000}{2} - \frac{90000}{2}\right) = 0.01 \left(500000 - 45000\right) = 0.01 \times 455000 = 4550 \]

### Step 5: Combine the results

Now, combine the results of the two integrals:

\[ \Delta H = 3 \left( 16100 + 4550 \right) = 3 \times 20550 = 61650 \, J \]

### Conclusion

The value of ΔH will be close to **61650 J** or **61.65 kJ**.

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