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Entropy is a measure of randomess of sys...

Entropy is a measure of randomess of system. When a liquid is converted to the vapour state entropy of the system increases. Entropy in the phase transformation is calculated using `Delta S = (Delta H)/(T)` but in reversible adiabatic process `Delta`S will be zero. The rise in temperature in isobaric or isochoric process increases the randomness of system, which is given by
`Delta S = "2.303 n C log"((T_(2))/(T_(1)))`
`C = C_(P) or C_(V)`
The temperature at whicgh liquid `H_(2)O`will be in equrilibrium with its vapour is (`Delta H and Delta S` for vapourisation are 50 kJ `mol^(-1)` and 0.15 kJ `mol^(-1)K^(-1)`)

A

`77.30^(@)C`

B

`60.33^(@)C`

C

`50^(@)C`

D

`100^(@)C`

Text Solution

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The correct Answer is:
To solve the problem, we need to find the temperature at which liquid H₂O will be in equilibrium with its vapor using the given values for the enthalpy change (ΔH) and the entropy change (ΔS) for vaporization. ### Step-by-Step Solution: 1. **Identify the Given Values**: - ΔH (enthalpy change for vaporization) = 50 kJ/mol - ΔS (entropy change for vaporization) = 0.15 kJ/(mol·K) 2. **Use the Formula for Temperature**: - The relationship between entropy change and enthalpy change during a phase transition is given by: \[ \Delta S = \frac{\Delta H}{T} \] - Rearranging the formula to solve for temperature (T): \[ T = \frac{\Delta H}{\Delta S} \] 3. **Substitute the Values**: - Substitute the values of ΔH and ΔS into the equation: \[ T = \frac{50 \text{ kJ/mol}}{0.15 \text{ kJ/(mol·K)}} \] 4. **Calculate the Temperature**: - Performing the calculation: \[ T = \frac{50}{0.15} = 333.33 \text{ K} \] 5. **Convert Temperature to Celsius**: - To convert from Kelvin to Celsius, use the formula: \[ T(°C) = T(K) - 273.15 \] - Thus: \[ T(°C) = 333.33 - 273.15 = 60.18 \text{ °C} \] - Rounding this value gives approximately: \[ T(°C) \approx 60.33 \text{ °C} \] 6. **Select the Correct Option**: - The calculated temperature of 60.33 °C matches option B. ### Final Answer: The temperature at which liquid H₂O will be in equilibrium with its vapor is approximately **60.33 °C** (Option B).
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