Phase transitions are ubiquitous in nature. We are all familiar with the different phase of water (vapour, liquid and ice) and with the change from one to another, the change of phase are called phase transitions. There are six ways a substance can change between these three phase, melting, freezing, evaporating, condensing sublimation and decomposition.
At `1 atm` pressure vaporisation of `1` mole of water from liquid `(75^(@)C)` to vapour `(120^(@)C)`.
`C_(v)(H_(2)O,l)=75 J "mole"^(-1)K^(-1), C_(p)(H_(2)O,g)=33.3J"mole"^(-1)K^(_1)`
`Delta H_(vap)` at `100^(@)C=40.7KJ//"mole"`
Calculate change in internal energy when
Water vapour at `100^(@)C` to `120^(@)C` ?

Options

A. `666 J`
B. `2998 J`
C. `499.72 J`
D. `166.28 J`

Answer

To calculate the change in internal energy (ΔU) when water vapor changes from 100°C to 120°C, we can follow these steps:

### Step 1: Identify the relevant parameters
We are given:
- Number of moles (n) = 1 mole
- Specific heat capacity of water vapor at constant pressure (C_p) = 33.3 J mole⁻¹ K⁻¹
- Initial temperature (T_initial) = 100°C
- Final temperature (T_final) = 120°C

### Step 2: Convert temperatures to Kelvin (if necessary)
For the change in temperature, we can use the Celsius scale directly since we are calculating a difference:
- ΔT = T_final - T_initial = 120°C - 100°C = 20°C

### Step 3: Calculate the change in internal energy (ΔU)
The change in internal energy for a process at constant pressure can be calculated using the formula:
\[ \Delta U = n \cdot C_p \cdot \Delta T \]
Substituting the values:
- n = 1 mole
- C_p = 33.3 J mole⁻¹ K⁻¹
- ΔT = 20 K (or °C, since the difference is the same)

Now, substituting these values into the formula:
\[ \Delta U = 1 \cdot 33.3 \, \text{J mole}^{-1} \text{K}^{-1} \cdot 20 \, \text{K} \]
\[ \Delta U = 666 \, \text{J} \]

### Step 4: Final Answer
Thus, the change in internal energy when water vapor changes from 100°C to 120°C is:
\[ \Delta U = 666 \, \text{J} \]

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