Methane ( Considered to be an ideal gas ) initially at `25^(@)C` and 1 bar pressure is heated at constant
pressure until the volume has dobbled . The variation of the molar heat capacity with absolute temperature
is given by `C_(P)=20 + 0.001 T`
where `C_(P)` is in `JK^(-1) mol^(-1)` . Calculate molar (a) `DeltaH`(b) `DeltaU`.

Explanation

To solve the problem, we will follow these steps:

### Given Data:
- Initial Temperature, \( T_1 = 25^\circ C = 298 \, K \)
- Initial Pressure, \( P = 1 \, bar \)
- Final Volume, \( V_2 = 2V_1 \) (Volume doubles)
- Molar Heat Capacity Variation: \( C_P = 20 + 0.001 T \) (in \( J \, K^{-1} \, mol^{-1} \))

### Step 1: Find the Final Temperature
Since the process occurs at constant pressure, we can use the relationship between volume and temperature for an ideal gas:

\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]

Substituting \( V_2 = 2V_1 \):

\[ \frac{V_1}{T_1} = \frac{2V_1}{T_2} \]

This simplifies to:

\[ T_2 = 2T_1 \]

Substituting \( T_1 = 298 \, K \):

\[ T_2 = 2 \times 298 = 596 \, K \]

### Step 2: Calculate the Change in Enthalpy (\( \Delta H \))
The change in enthalpy at constant pressure is given by:

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

Substituting the expression for \( C_P \):

\[ \Delta H = \int_{298}^{596} (20 + 0.001 T) \, dT \]

Calculating the integral:

\[ \Delta H = \left[ 20T + 0.0005T^2 \right]_{298}^{596} \]

Calculating the values at the limits:

1. For \( T = 596 \):

\[ 20 \times 596 + 0.0005 \times (596^2) = 11920 + 0.0005 \times 355216 = 11920 + 177.608 = 12109.608 \]

2. For \( T = 298 \):

\[ 20 \times 298 + 0.0005 \times (298^2) = 5960 + 0.0005 \times 88804 = 5960 + 44.002 = 6004.002 \]

Now, substituting back into the equation for \( \Delta H \):

\[ \Delta H = 12109.608 - 6004.002 = 6105.606 \, J/mol \]

### Step 3: Calculate the Change in Internal Energy (\( \Delta U \))
Using the relation between enthalpy and internal energy:

\[ \Delta H = \Delta U + R \Delta T \]

Where \( R = 8.314 \, J \, K^{-1} \, mol^{-1} \) and \( \Delta T = T_2 - T_1 = 596 - 298 = 298 \, K \).

Calculating \( R \Delta T \):

\[ R \Delta T = 8.314 \times 298 = 2477.572 \, J/mol \]

Now substituting into the equation for \( \Delta U \):

\[ \Delta U = \Delta H - R \Delta T = 6105.606 - 2477.572 = 3628.034 \, J/mol \]

### Final Answers:
- (a) \( \Delta H = 6105.606 \, J/mol \) or \( 6.106 \, kJ/mol \)
- (b) \( \Delta U = 3628.034 \, J/mol \) or \( 3.628 \, kJ/mol \)