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Chapter 6: Problem 131

Calculate \(w\) and \(\Delta E\) when 1 mole of a liquid is vaporized at its boiling point \(\left(80 .^{\circ} \mathrm{C}\right)\) and 1.00 atm pressure. \(\Delta H_{\text { vap }}\) for the liquid is 30.7 \(\mathrm{kJ} / \mathrm{mol}\) at \(80 .^{\circ} \mathrm{C} .\)

Short Answer

Expert verified
The work done during the vaporization process is \(w = -2937.7 \, \text{J}\) and the change in internal energy is \(\Delta E = 27762.3 \, \text{J/mol}\).

Step by step solution

01

Calculate work done by the system

We are considering an isothermal process, so the temperature is constant. The formula for work done during constant temperature and pressure process on 1 mole of liquid is: \(w = - P\Delta V\) where: - P is the external pressure, and - ΔV is the change in volume during the vaporization process. The negative sign indicates that the work is done by the system, which means it's an expansion process against the surrounding pressure. However, we don't know the ΔV to determine the work. So, we need to find the change in volume during the vaporization process.
02

Calculate the change in volume during vaporization

We know that the boiling point of the liquid is 80°C, and it is vaporized under atmospheric pressure (1.00 atm). We are assuming that the vapor acts as an ideal gas. From the ideal gas equation, we can find the molar volume of the vapor: \(PV = nRT\) where: - n is the number of moles (in this case 1 mole), - R is the gas constant (0.0821 L atm/mol K, since the pressure is given in atm), and - T is the temperature in Kelvin (353.15 K). Solving for V: \(V = \frac{nRT}{P}\) Since we know that the molar volume of condensed liquid is negligible compared to molar volume of vapor, we can approximate the change in volume (ΔV) with the molar volume of the vapor (V): \(\Delta V \approx V\)
03

Plug in the values and calculate work done by the system

Now, we can plug in the values to calculate the work done during the vaporization process: \(w = - P\Delta V = -P\left(\frac{nRT}{P}\right)\) \(w = -nRT\) With n = 1, R = 0.0821 L atm/mol K, and T = 353.15 K: \(w = -(1)(0.0821)(353.15) = -29.0 \, \text{L atm}\) To convert the work to Joules multiply with 101.3 J/L atm: \(w = -29.0 \, \text{L atm} \times 101.3 \, \frac{\text{J}}{\text{L atm}} = -2937.7 \, \text{J}\) Now, we will calculate the change in internal energy (ΔE) for the vaporization process.
04

Use the first law of thermodynamics to calculate the change in internal energy

The first law of thermodynamics relates the heat absorbed by the system (Q), the work done by the system (w), and the change in internal energy (ΔE): \(\Delta E = Q + w\) We know that the enthalpy of vaporization (ΔH_vap) is 30.7 kJ/mol, and since the process is occurring at constant pressure, we have: \(Q = \Delta H_{\text { vap }}\) Convert ΔH_vap to Joules: ΔH_vap = (30.7 kJ/mol) × (1000 J/1 kJ) = 30700 J/mol Now plug in the values for Q and w to find ΔE: \(\Delta E = Q + w = 30700 - 2937.7 = 27762.3 \, \text{J/mol}\) Our final results are: \(w = -2937.7 \, \text{J}\) \(\Delta E = 27762.3 \, \text{J/mol}\)

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Most popular questions from this chapter

Consider 2.00 moles of an ideal gas that are taken from state \(A\) \(\left(P_{A}=2.00 \mathrm{atm}, V_{A}=10.0 \mathrm{L}\right)\) to state \(B\left(P_{B}=1.00 \mathrm{atm}, V_{B}=\right.\) 30.0 \(\mathrm{L}\) ) by two different pathways: These pathways are summarized on the following graph of \(P\) versus \(V :\) Calculate the work (in units of \(\mathrm{J} )\) associated with the two path- ways. Is work a state function? Explain.

Using the following data, calculate the standard heat of formation of ICl \((g)\) in \(\mathrm{kJ} / \mathrm{mol} :\) $$\begin{array}{ll}{\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{Cl}(g)} & {\Delta H^{\circ}=242.3 \mathrm{kJ}} \\ {\mathrm{I}_{2}(g) \longrightarrow 2 \mathrm{I}(g)} & {\Delta H^{\circ}=151.0 \mathrm{kJ}} \\ {\mathrm{ICl}(g) \longrightarrow \mathrm{I}(g)+\mathrm{Cl}(g)} & {\Delta H^{\circ}=211.3 \mathrm{kJ}} \\ {\mathrm{I}_{2}(s) \longrightarrow \mathrm{I}_{2}(g)} & {\Delta H^{\circ}=62.8 \mathrm{kJ}}\end{array}$$

In a coffee-cup calorimeter, 50.0 \(\mathrm{mL}\) of 0.100\(M \mathrm{AgNO}_{3}\) and 50.0 \(\mathrm{mL}\) of 0.100 \(\mathrm{M} \mathrm{HCl}\) are mixed to yield the following reaction: $$\mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{AgCl}(s)$$ The two solutions were initially at \(22.60^{\circ} \mathrm{C}\) , and the final temperature is \(23.40^{\circ} \mathrm{C}\) Calculate the heat that accompanies this reacture in kJ/mol of AgCl formed. Assume that the combined solution has a mass of 100.0 \(\mathrm{g}\) and a specific heat capacity of 4.18 \(\mathrm{J} / \rho \mathrm{C} \cdot \mathrm{g} .\)

Consider the dissolution of \(\mathrm{CaCl}_{2} :\) $$ \mathrm{CaCl}_{2}(s) \longrightarrow \mathrm{Ca}^{2+}(a q)+2 \mathrm{Cl}^{-}(a q) \quad \Delta H=-81.5 \mathrm{kJ} $$ An 11.0 -g sample of \(\mathrm{CaCl}_{2}\) is dissolved in 125 g water, with both substances at \(25.0^{\circ} \mathrm{C}\) . Calculate the final temperature of the solution assuming no heat loss to the surroundings and assuming the solution has a specific heat capacity of 4.18 $\mathrm{J} /^{\prime} \mathrm{C} \cdot \mathrm{g} .$

A sample of an ideal gas at 15.0 atm and 10.0 \(\mathrm{L}\) is allowed to expand against a constant external pressure of 2.00 atm at a constant temperature. Calculate the work in units of kJ for the gas expansion. (Hint: Boyle's law applies.)
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