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Chapter 10: Problem 131

When 1 mole of benzene is vaporized at a constant pressure of \(1.00\) atm and at its boiling point of \(353.0 \mathrm{~K}, 30.79 \mathrm{~kJ}\) of energy (heat) is absorbed and the volume change is \(+28.90 \mathrm{~L}\). What are \(\Delta E\) and \(\Delta H\) for this process?

Short Answer

Expert verified
For this process, the change in internal energy (ΔE) is \(1.508 \mathrm{~kJ}\), and the change in enthalpy (ΔH) is \(30.79 \mathrm{~kJ}\).

Step by step solution

01

Calculate the work done on the system

We can use the formula for work, which is w = -PΔV. Since we are given the pressure P = 1.00 atm and the volume change ΔV = +28.90 L, we can plug these values into the formula: w = -PΔV = -(1.00 atm)(28.90 L) To continue, we need to convert pressure and volume from atm and L, respectively, to Joules (J). We can use the conversion factor: 1 L·atm = 101.325 J. So, we have: w = -(1.00 × 28.90 × 101.325 J) = -29282.025 J ≈ -29.282 kJ
02

Determine the change in internal energy ΔE

According to the first law of thermodynamics, ΔE = q + w. We are given the heat absorbed (q) as 30.79 kJ, and we calculated work (w) as -29.282 kJ. So, we can now calculate ΔE: ΔE = q + w = 30.79 kJ - 29.282 kJ = 1.508 kJ
03

Determine the change in enthalpy ΔH

To calculate the change in enthalpy ΔH, we use the formula ΔH = ΔE + PΔV, where we already have ΔE = 1.508 kJ, P = 1.00 atm, and ΔV = 28.90 L: ΔH = ΔE + PΔV = 1.508 kJ + (1.00 atm × 28.90 L) Now, we need to convert (1.00 atm × 28.90 L) to Joules using the conversion factor 1 L·atm = 101.325 J: ΔH = 1.508 kJ + (1.00 × 28.90 × 101.325 J) = 1.508 kJ + 29282.025 J = 1.508 kJ + 29.282 kJ = 30.79 kJ So, for this process: - ΔE = 1.508 kJ - ΔH = 30.79 kJ

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