Question

You have an ideal gas with an initial volume of 1.0 L and initial pressure of 10 atm. You decide to change the conditionssuch that Pfinal = 20 atm and Vfinal = 20 L To make things more interesting, you and a friend each carry out this changein two steps. You first change the volume so that at one point P = 10 atm and V = 20 atm. Your friend first changes thepressure so that at one point P = 20 atm and V = 1.0 atm. Both of you, of course, end up with the same final conditionsafter the second step.a. How should your AE's. AH's, q‘s, and w's compare with those of your friend? Why?b. Calculate AE, AH. q, and w for each case. Do they make sense? Explain.

Answer 1

Answer and Explanation:

The change of internal energy (ΔE) and the change of enthalpy (ΔH) are defined as state functions thus, they only depend of the initial and final state. Since they do not depend on the path by which the system arrived, both experiments will have the same ΔE and ΔH.

On the other hand, total heat (q) and total work (w) are not state functions thus, they depend on the path by which the system arrived. Since the two path are different the result will be different.

In this experiment, there is a process at constant temperature. For ideal gases, all reversible changes at constant temperature has no change in ΔE and ΔH. Therefore, for both experiments ΔE=0 and ΔH=0.

Thermodynamics defines ΔE as the sum between heat and work, since this is cero, q = -w. Also, the definition of work is W = -P.dV

For the first experiment, the work of the first and second steps are

W1 = -10atm x (20L - 1L) = -190atm.L and W2 = 0

For the second experiment, the work of the first and second steps are

W1 = 0 and W2 = -10atm x (20L - 1L) = -190atm.L

Therefore, the total work between the two steps are for both -190atm.L and the total heat for both is 190atm.L.

The result of heat and work do not make sense because both increase at the end and this two variables are inversely proportional. These experiments can not be made at constant temperature.