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Phase transitions are ubiquitous in natu...

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 liquid at `100^(@)C` to vapour at `100^(@)C` ?

A

`40700 J`

B

`37598.878 J`

C

`43801.1J`

D

`31012.J`

Text Solution

AI Generated Solution

The correct Answer is:
To calculate the change in internal energy (ΔU) when water changes from liquid at 100°C to vapor at 100°C, we can use the relationship between the change in internal energy and the enthalpy of vaporization (ΔH_vap). Here are the steps to solve the problem: ### Step 1: Understand the Phase Change We are considering the phase transition of water from liquid to vapor at the same temperature (100°C). During this process, the temperature remains constant, and we can use the enthalpy of vaporization to find the change in internal energy. ### Step 2: Use the Formula for Change in Internal Energy The change in internal energy (ΔU) during a phase change at constant temperature can be expressed as: \[ \Delta U = n \cdot \Delta H_{vap} \] where: - \( n \) = number of moles of water (1 mole in this case) - \( \Delta H_{vap} \) = enthalpy of vaporization at 100°C (given as 40.7 kJ/mol) ### Step 3: Substitute the Values Given: - \( n = 1 \, \text{mole} \) - \( \Delta H_{vap} = 40.7 \, \text{kJ/mol} = 40700 \, \text{J/mol} \) (since 1 kJ = 1000 J) Now, substituting these values into the formula: \[ \Delta U = 1 \, \text{mol} \cdot 40700 \, \text{J/mol} \] ### Step 4: Calculate ΔU Calculating the above expression gives: \[ \Delta U = 40700 \, \text{J} \] ### Conclusion Thus, the change in internal energy when water changes from liquid to vapor at 100°C is: \[ \Delta U = 40700 \, \text{J} \]
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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 liquid at 75^(@)C to 100^(@)C ?

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 ?

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