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Act like a helpful tutor and exlain me :© 1. Calculate the values of AQ. AW, AU. and AH required 10 take one mole of solid ethanol from its normal freezing point 10 one mole of cthanol vapor at its normal boiling point (8) Look up the required thermophy sical data and make the calculation by hand.

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Act like a helpful tutor and exlain me :Uploaded Image© 1. Calculate the values of AQ. AW, AU. and AH required 10 take one mole of solid ethanol from its normal freezing point 10 one mole of cthanol vapor at its normal boiling point (8) Look up the required thermophy sical data and make the calculation by hand.

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Asked by: Alok Kumar
Created at: 2026-01-29 09:27:54

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Certainly! Let's break down the problem step by step. --- ## **Problem Restatement** **Calculate the values of \( \Delta Q \), \( \Delta W \), \( \Delta U \), and \( \Delta H \) required to take one mole of solid ethanol from its normal freezing point to one mole of ethanol vapor at its normal boiling point.** --- ## **Step 1: Understand the Process** We're taking **1 mole of solid ethanol**: - From its **normal freezing point** (solid ethanol at its melting temperature) - To **ethanol vapor** at its **normal boiling point** This involves several steps: 1. **Heating solid ethanol** from its freezing point to its melting point (if needed) 2. **Melting (fusion)**: Solid \( \rightarrow \) Liquid 3. **Heating liquid ethanol** from melting point to boiling point 4. **Vaporization**: Liquid \( \rightarrow \) Vapor --- ## **Step 2: Required Thermophysical Data** We'll need: - **Freezing point (melting point)** of ethanol: \( T_f = 159 \, \text{K} \) - **Boiling point** of ethanol: \( T_b = 351 \, \text{K} \) - **Molar heat capacity of solid ethanol (\(C_{p,s}\))**: ~\( 112 \, \text{J/mol·K} \) - **Molar heat capacity of liquid ethanol (\(C_{p,l}\))**: ~\( 112 \, \text{J/mol·K} \) - **Enthalpy of fusion (\( \Delta H_{fus} \))**: \( 4.9 \, \text{kJ/mol} \) - **Enthalpy of vaporization (\( \Delta H_{vap} \))**: \( 38.6 \, \text{kJ/mol} \) --- ## **Step 3: Calculating \( \Delta Q \) (Total Heat)** The **total heat required (\( \Delta Q \))** is the sum of: 1. Heat to melt the solid: \( \Delta H_{fus} \) 2. Heat to warm the liquid from \( T_f \) to \( T_b \): \( nC_{p,l}(T_b - T_f) \) 3. Heat to vaporize: \( \Delta H_{vap} \) So, \[ \Delta Q = \Delta H_{fus} + nC_{p,l}(T_b - T_f) + \Delta H_{vap} \] Where \( n = 1 \) mole. --- ## **Step 4: Calculating \( \Delta H \) (Enthalpy Change)** For this process at constant pressure (assuming atmospheric), the total heat for a reversible process equals the enthalpy change: \[ \Delta H = \Delta Q \] --- ## **Step 5: Calculating \( \Delta U \) (Internal Energy Change)** \[ \Delta U = \Delta H - \Delta (PV) \] For ideal gases, \( \Delta (PV) = nR(T_{final} - T_{initial}) \) - \( T_{final} = \) boiling point in K = 351 K - \( T_{initial} = \) freezing point in K = 159 K --- ## **Step 6: Calculating \( \Delta W \) (Work Done)** At constant pressure, the work done by the system is: \[ \Delta W = P \Delta V \] - Most of the volume change occurs during vaporization. --- ## **Step 7: Plug in Values** ### **Data:** - \( T_f = 159 \) K - \( T_b = 351 \) K - \( C_{p,l} = 112 \) J/mol·K - \( \Delta H_{fus} = 4.9 \) kJ/mol = 490 J/mol - \( \Delta H_{vap} = 38.6 \) kJ/mol = 38600 J/mol - \( R = 8.314 \) J/mol·K ### **Calculations:** #### **1. Heating Liquid Ethanol:** \[ q_1 = n C_{p,l} (T_b - T_f) = 1 \times 112 \times (351 - 159) = 1 \times 112 \times 192 = 21,504 \text{ J} \] #### **2. Melting:** \[ q_2 = \Delta H_{fus} = 4,900 \text{ J} \] #### **3. Vaporization:** \[ q_3 = \Delta H_{vap} = 38,600 \text{ J} \] #### **Total Heat (\( \Delta Q \) and \( \Delta H \)):** \[ \Delta Q = \Delta H = 4,900 + 21,504 + 38,600 = 65,004 \text{ J} \] #### **4. Internal Energy (\( \Delta U \)):** \[ \Delta U = \Delta H - nR(T_b - T_f) \] \[ = 65,004 - (1 \times 8.314 \times (351 - 159)) \] \[ = 65,004 - (8.314 \times 192) \] \[ = 65,004 - 1,596.29 = 63,407.71 \text{ J} \] #### **5. Work Done (\( \Delta W \)):** Work is mainly from liquid to vapor (since solids and liquids have much smaller volume changes): \[ \Delta W = P \Delta V \approx nRT_b \] \[ = 1 \times 8.314 \times 351 = 2,918 \text{ J} \] --- ## **Summary Table** | Quantity | Value (J) | |--------------------|-------------| | \( \Delta Q \) | 65,004 | | \( \Delta H \) | 65,004 | | \( \Delta U \) | 63,408 | | \( \Delta W \) | 2,918 | --- ### **Key Points** - **\( \Delta Q \)** and **\( \Delta H \)** are practically the same at constant pressure. - **\( \Delta U \)** is slightly less due to work done against the atmosphere. - **\( \Delta W \)** comes primarily from vaporization. --- #### If you'd like, I can show the step-by-step math for any part, or help you look up any of the data! Let me know what you need more details on.

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