One moles of an ideal gas which `C_(V) = 3//2 R` is heated at a constant pressure of `1 atm` from `25^(@)C` to `100^(@)C`. Calculate `DeltaU, DeltaH` and the entropy change during the process.

Explanation

To solve the problem step by step, we will calculate the change in internal energy (ΔU), change in enthalpy (ΔH), and the change in entropy (ΔS) for one mole of an ideal gas heated at constant pressure.

### Step 1: Calculate ΔU (Change in Internal Energy)

The formula for the change in internal energy (ΔU) is given by:
\[ \Delta U = n C_V \Delta T \]
Where:
- \( n = 1 \) mole (given)
- \( C_V = \frac{3}{2} R \)
- \( \Delta T = T_2 - T_1 = (100 + 273) - (25 + 273) = 100 - 25 = 75 \, \text{°C} \)

Now substituting the values:
\[ \Delta U = 1 \times \left(\frac{3}{2} R\right) \times 75 \]
Using \( R = 2 \, \text{cal/K·mol} \):
\[ \Delta U = 1 \times \left(\frac{3}{2} \times 2\right) \times 75 = 3 \times 75 = 225 \, \text{cal} \]

### Step 2: Calculate ΔH (Change in Enthalpy)

The formula for the change in enthalpy (ΔH) is given by:
\[ \Delta H = n C_P \Delta T \]
Where:
- \( C_P = C_V + R = \frac{3}{2} R + R = \frac{5}{2} R \)

Now substituting the values:
\[ \Delta H = 1 \times \left(\frac{5}{2} R\right) \times 75 \]
Using \( R = 2 \, \text{cal/K·mol} \):
\[ \Delta H = 1 \times \left(\frac{5}{2} \times 2\right) \times 75 = 5 \times 75 = 375 \, \text{cal} \]

### Step 3: Calculate ΔS (Change in Entropy)

The formula for the change in entropy (ΔS) at constant pressure is given by:
\[ \Delta S = n C_P \ln\left(\frac{T_2}{T_1}\right) \]
Where:
- \( T_1 = 25 + 273 = 298 \, \text{K} \)
- \( T_2 = 100 + 273 = 373 \, \text{K} \)

Now substituting the values:
\[ \Delta S = 1 \times \left(\frac{5}{2} R\right) \ln\left(\frac{373}{298}\right) \]
Using \( R = 2 \, \text{cal/K·mol} \):
\[ \Delta S = 1 \times \left(\frac{5}{2} \times 2\right) \ln\left(\frac{373}{298}\right) = 5 \ln\left(\frac{373}{298}\right) \]
Calculating \( \ln\left(\frac{373}{298}\right) \):
\[ \ln\left(\frac{373}{298}\right) \approx 0.251 \]
Thus,
\[ \Delta S \approx 5 \times 0.251 \approx 1.255 \, \text{cal/K·mol} \]

### Summary of Results:
- \( \Delta U = 225 \, \text{cal} \)
- \( \Delta H = 375 \, \text{cal} \)
- \( \Delta S \approx 1.255 \, \text{cal/K·mol} \)