Answered: A 1 mole sample of ideal gas expands…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Question

A 1 mole sample of ideal gas expands isothermally at 300 K from a volume of 20.00 L to a volume of 30.00 L. Calculate the ΔG and the ΔS for this process.

Expert Solution

Step 1

Since the process is isothermal, the change in temperature is equal to zero.

The change in internal energy (∆U) is equal to zero.

The change in enthalpy (∆H) is equal to ∆U + nRT where n is the number of moles of gas, R is the universal gas constant and T is the temperature. The value of ∆H = 1 × 8.314 × 300 = 2494.2 joules. ∆U = q + W. Since ∆U = 0, q = -W. The change in entropy (∆S) = q/T. The change in entropy for this reversible isothermal expansion of gas can be determined as follows:

$\begin{array}{rcl} ∆ S & = & 2.303 nRlog (\frac{V_{2}}{V_{1}}) \\ = & 2.303 \times 1 \times 8.314 \log (\frac{30}{20}) \\ = & 19.15 \times 0.18 \\ = & 3.45 J / K \end{array}$

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