A new flourocarbon of molar mass 102 g `mol^(-1)` was placed in an electricity heated vessel. Whwn the pressure was 650 torr, the liquid boiled at `77^(@)` C. After the boiling point had been reached , it was found that a current of 0.25 A from a 12.0 volt supply passed for 600 sec vaporises 1.8g of the sample. The molar enthalpy and enternal energy of vaporisation of new flourocarbon will be :

Options

A. `DeltaH=102KJ//mol, DeltaE=99.1KJ//mol`
B. `DeltaH=95KJ//mol, DeltaE=100.3 KJ//mol`
C. `DeltaH=107 KJ//mol, DeltaE=105.1 KJ//mol`
D. `DeltaH=92.7 KJ//mol, DeltaE=97.4 KJ//mol`

Answer

To solve the problem of finding the molar enthalpy and internal energy of vaporization of the new fluorocarbon, we will follow these steps:

### Step 1: Calculate the Energy Supplied
We know that power (P) can be calculated using the formula:
\[ P = V \times I \]
where \( V \) is the voltage and \( I \) is the current.

Given:
- Voltage, \( V = 12.0 \, \text{V} \)
- Current, \( I = 0.25 \, \text{A} \)

Calculating the power:
\[ P = 12.0 \, \text{V} \times 0.25 \, \text{A} = 3.0 \, \text{W} \]

Now, we can calculate the energy (E) supplied over time (t):
\[ E = P \times t \]
where \( t = 600 \, \text{s} \).

Calculating the energy:
\[ E = 3.0 \, \text{W} \times 600 \, \text{s} = 1800 \, \text{J} \]

### Step 2: Calculate the Number of Moles of the Sample Vaporized
To find the number of moles (n) of the fluorocarbon that was vaporized, we use the formula:
\[ n = \frac{\text{mass}}{\text{molar mass}} \]

Given:
- Mass of the sample vaporized = 1.8 g
- Molar mass = 102 g/mol

Calculating the number of moles:
\[ n = \frac{1.8 \, \text{g}}{102 \, \text{g/mol}} = 0.01765 \, \text{mol} \]

### Step 3: Calculate the Molar Enthalpy of Vaporization
The molar enthalpy of vaporization (\( \Delta H \)) can be calculated using the formula:
\[ \Delta H = \frac{E}{n} \]

Substituting the values:
\[ \Delta H = \frac{1800 \, \text{J}}{0.01765 \, \text{mol}} = 102,000 \, \text{J/mol} \]
Converting to kilojoules:
\[ \Delta H = 102 \, \text{kJ/mol} \]

### Step 4: Calculate the Internal Energy of Vaporization
We can use the relationship between enthalpy and internal energy:
\[ \Delta H = \Delta U + \Delta n RT \]

Where:
- \( \Delta n \) is the change in the number of moles of gas (for vaporization, \( \Delta n = 1 \))
- \( R \) is the universal gas constant \( = 8.314 \, \text{J/(mol K)} \)
- \( T \) is the temperature in Kelvin

Convert the boiling point from Celsius to Kelvin:
\[ T = 77 \, \text{C} + 273 = 350 \, \text{K} \]

Now substituting into the equation:
\[ \Delta H = \Delta U + (1)(8.314 \, \text{J/(mol K)})(350 \, \text{K}) \]

Calculating \( \Delta n RT \):
\[ \Delta n RT = 8.314 \times 350 = 2909.9 \, \text{J} \]
Converting to kilojoules:
\[ \Delta n RT = 2.9099 \, \text{kJ} \]

Now substituting back to find \( \Delta U \):
\[ 102 \, \text{kJ/mol} = \Delta U + 2.9099 \, \text{kJ} \]
\[ \Delta U = 102 - 2.9099 = 99.0901 \, \text{kJ/mol} \]

### Final Results
- Molar Enthalpy of Vaporization, \( \Delta H = 102 \, \text{kJ/mol} \)
- Internal Energy of Vaporization, \( \Delta U \approx 99.1 \, \text{kJ/mol} \)