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Quantities such as the isothermal compressibility help us to determine changes in the thermodynamic state functions for "real substances". These quantities can be measured through direct laboratory measurements, or through the use of substance-specific equations of state, such as the van der Waals equation. P = RT a Vm-b Vm² (a) Work through the math to obtain a formula for the isothermal compressibility of a van der Waals gas. KT = - 1 (Vm) Vm ӘР T

Quantities such as the isothermal compressibility help us to determine changes in the thermodynamic state functions for "real substances". These quantities can be measured through direct laboratory measurements, or through the use of substance-specific equations of state, such as the van der Waals equation. P = RT a Vm-b Vm² (a) Work through the math to obtain a formula for the isothermal compressibility of a van der Waals gas. KT = - 1 (Vm) Vm ӘР T

Chemistry
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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I need this to be done with steps a and b included, for a I got KT=(-1/Vm)*((2a/Vm^3)-(RT)/(Vm-b)2)-1 NOTICE THAT THIS IS A HUGE RECIPRICAL, if you do the algebra you should get:

Vm2(Vm-b)2
____________

2a(Vm-b)2-RTVm3

If that helps but in your walkthrough of part C please explain how RT will ever disappear, unless I am just missing a connection, even when you walk it through from the full partial of Vm wrt pressure at constant temperature...

3. Quantities such as the isothermal compressibility help us to determine changes in the thermodynamic state functions for "real substances". These quantities can be measured through direct laboratory measurements, or through the use of substance-specific equations of state, such as the van der Waals equation. RT a 2 Vm Vm² (a) Work through the math to obtain a formula for the isothermal compressibility of a van der Waals gas. P = KT == - b 1 /0Vm Vm Op T [Hint: this will probably be significantly easier if you use the reciprocal relationship for partial derivatives! This way, you can work from the van der Waals equation as given above, with P as the dependent variable. The resulting derivative is not necessarily “simple”, but you should be able to navigate through with some judicious use of the chain rule, and a bit of algebraic maneuvering.] (b) Once you have obtained your result, demonstrate that the units work out correctly. (c) Next, demonstrate that your expression reduces to the ideal gas result (KT = 1) at large values of Vm. (d) Calculate the value of KT for a hypothetical ideal gas at T = 525 K and Vm = 2.50 L mol-¹, and again at T = 525 K and Vm = 25.0 L mol-¹. (e) Calculate the value of KT for ethanol (as a van der Waals gas, use parameters from E & R table 7.4) when T = 525 K and Vm = 2.50 L mol-¹, and again at T = 525 K and Vm = 25.0 L mol-¹.
Transcribed Image Text:3. Quantities such as the isothermal compressibility help us to determine changes in the thermodynamic state functions for "real substances". These quantities can be measured through direct laboratory measurements, or through the use of substance-specific equations of state, such as the van der Waals equation. RT a 2 Vm Vm² (a) Work through the math to obtain a formula for the isothermal compressibility of a van der Waals gas. P = KT == - b 1 /0Vm Vm Op T [Hint: this will probably be significantly easier if you use the reciprocal relationship for partial derivatives! This way, you can work from the van der Waals equation as given above, with P as the dependent variable. The resulting derivative is not necessarily “simple”, but you should be able to navigate through with some judicious use of the chain rule, and a bit of algebraic maneuvering.] (b) Once you have obtained your result, demonstrate that the units work out correctly. (c) Next, demonstrate that your expression reduces to the ideal gas result (KT = 1) at large values of Vm. (d) Calculate the value of KT for a hypothetical ideal gas at T = 525 K and Vm = 2.50 L mol-¹, and again at T = 525 K and Vm = 25.0 L mol-¹. (e) Calculate the value of KT for ethanol (as a van der Waals gas, use parameters from E & R table 7.4) when T = 525 K and Vm = 2.50 L mol-¹, and again at T = 525 K and Vm = 25.0 L mol-¹.
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