What is the value of change in internal energy at 1 atm in the process?
`H_(2)O(l,323K)rarrH_(2)O(g,423K)`
Given : `C_(v,m)(H_(2)O,l)=75.0JK^(-1) mol^(-1):" "C_(p,m)(H_(2)O,g)=33.314JK^(-1)mol^(-1)`
`DeltaH_(vap)"ar 373 K"=40.7KJ//mol`
(a)`42.91kJ//mol`
(b)`43086kJ//mol`
(c)`42.6kJ//mol`
(d)`49.6kJ//mol`

Options

A. `42.91kJ//mol`
B. `43086kJ//mol`
C. `42.6kJ//mol`
D. `49.6kJ//mol`

Explanation

To find the change in internal energy (ΔU) for the process \( H_2O(l, 323K) \rightarrow H_2O(g, 423K) \), we will break it down into three steps and calculate the internal energy change for each step.

### Step 1: Heating liquid water from 323 K to 373 K
The change in internal energy for this step can be calculated using the formula:
\[ \Delta U_1 = C_{v,m} \cdot \Delta T \]
Where:
- \( C_{v,m}(H_2O, l) = 75.0 \, J \, K^{-1} \, mol^{-1} \)
- \( \Delta T = T_2 - T_1 = 373 \, K - 323 \, K = 50 \, K \)

Now substituting the values:
\[ \Delta U_1 = 75.0 \, J \, K^{-1} \, mol^{-1} \cdot 50 \, K = 3750 \, J \, mol^{-1} = 3.75 \, kJ \, mol^{-1} \]

### Step 2: Vaporization at 373 K
The change in internal energy during the vaporization can be calculated using the heat of vaporization:
\[ \Delta U_2 = \Delta H_{vap} - R \cdot \Delta n \cdot T \]
Where:
- \( \Delta H_{vap} = 40.7 \, kJ \, mol^{-1} \)
- \( R = 8.314 \, J \, K^{-1} \, mol^{-1} = 0.008314 \, kJ \, K^{-1} \, mol^{-1} \)
- \( \Delta n = 1 \) (since 1 mole of liquid turns into 1 mole of gas)
- \( T = 373 \, K \)

Now substituting the values:
\[ \Delta U_2 = 40.7 \, kJ \, mol^{-1} - (0.008314 \, kJ \, K^{-1} \, mol^{-1} \cdot 1 \cdot 373 \, K) \]
Calculating the second term:
\[ 0.008314 \cdot 373 = 3.101 \, kJ \, mol^{-1} \]
Now substituting back:
\[ \Delta U_2 = 40.7 \, kJ \, mol^{-1} - 3.101 \, kJ \, mol^{-1} = 37.599 \, kJ \, mol^{-1} \]

### Step 3: Heating vapor from 373 K to 423 K
The change in internal energy for this step can be calculated similarly:
\[ \Delta U_3 = C_{p,m} \cdot \Delta T \]
Where:
- \( C_{p,m}(H_2O, g) = 33.314 \, J \, K^{-1} \, mol^{-1} \)
- \( \Delta T = 423 \, K - 373 \, K = 50 \, K \)

Now substituting the values:
\[ \Delta U_3 = 33.314 \, J \, K^{-1} \, mol^{-1} \cdot 50 \, K = 1665.7 \, J \, mol^{-1} = 1.6657 \, kJ \, mol^{-1} \]

### Total Change in Internal Energy
Now, we can sum up all the changes in internal energy:
\[ \Delta U_{total} = \Delta U_1 + \Delta U_2 + \Delta U_3 \]
Substituting the values:
\[ \Delta U_{total} = 3.75 \, kJ \, mol^{-1} + 37.599 \, kJ \, mol^{-1} + 1.6657 \, kJ \, mol^{-1} \]
Calculating:
\[ \Delta U_{total} = 3.75 + 37.599 + 1.6657 = 43.0147 \, kJ \, mol^{-1} \]

### Final Result
Rounding to two decimal places, we find:
\[ \Delta U_{total} \approx 42.6 \, kJ \, mol^{-1} \]

### Conclusion
The value of change in internal energy at 1 atm for the process \( H_2O(l, 323K) \rightarrow H_2O(g, 423K) \) is approximately \( 42.6 \, kJ \, mol^{-1} \).

### Answer
(c) \( 42.6 \, kJ \, mol^{-1} \)

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