--%>
Assignment Help - homework help

Problem on Adiabatic expansion

Calculate the change in entropy for the system for each of the following cases. Explain the sign that you obtain by a physical argument

a) A gas undergoes a reversible, adiabatic expansion from an initial state at 500 K, 1 MPa, and 8.314 L to a final volume of 16.628 L.

b) One mole of methane vapor is condensed at its boiling point, 111 K; Δhv = 8.2 [kJ/mol].

c) One mole of liquid water is cooled from 100°C to 0°C. Take the average heat capacity of water to be 4.2 JK-1g-1.

d) Two blocks of the same metal with equal mass are at different temperatures, 200°C and 100°C. These blocks are brought together and allowed to come to the same temperature. Assume that these blocks are isolated from their surroundings. The average heat capacity of the metal is 24 JK-1mol-1.

E

Expert

Verified

(a) Since the heat transfer, ΔQ = 0, in reversible adiabatic process, the entropy change,

ΔS = ΔQ/T = 0

(b) ΔS = Δhv/T = (-8.2 kJ/mol)/111 K = -0.074 kJ/(mol.K) = -74 J/(mol.K)

Since one mol is condensed, -74J/K is the entropy change, and this heat taken up by surrounding whose entropy change is positive 74J/K, and hence the entropy change of system plus surrounding is zero, in confirmation with the second law of thermodynamics.

(c) ΔS = ΔQ/T = ∫cp,avgdT/T = cp,avg ∫dT/T = cp,avg ln (T2/T1) = 4.2 ln(273/373) = = -1.31 J/(gK).

But we have 1 mol of water, i.e. 18 gm of water. Hence ΔS = -1.31 x 18 = -23.58 J/K

The negative sign implies that heat is lost or transferred from system to surrounding.

In other words water is cooled, by transferring the heat, hence the change in entropy is negative, while the surrounding gain the same amount of heat and for it the change in entropy is positive, hence the total change in entropy is zero, i.e. System + Surroundings.

(d) Let the equilibrium temperature be T,

mCp(200 – T) = mCp(T – 100)
(200 – T) = (T – 100)
T = 150oC

Total change in entropy of the system,

        = change in entropy of 1st block + change in entropy of 2nd block

        = cp ln (T2/T1) + cp ln (T2/T1)

        = 24ln (423/473) + 24ln (423/373)

        = 0.338 J/mol.K

Thus the entropy change is positive in this case, implying there are more configurations when the two blocks are allowed to interact.

   Related Questions in Chemistry

  • Q : Explain preparation and properties of

    It may be prepared by the action of phosphorus on thionyl chloride.P4 + 8SOCl2    4

    		•
  • Q : What are various structure based

    This classification of polymers is based upon how the monomeric units are linked together. Based on their structure, the polymers are classified as: 1. Linear polymers: these are the polymers in which monomeric units are linked together to form long straight c

    		•
  • Q : Problem on molecular weight of solid

    The vapor pressure of pure benzene at a certain temperature is 200 mm Hg. At the same temperature the vapor pressure of a solution containing 2g of non-volatile non-electrolyte solid in 78g of benzene is 195 mm Hg. What is the molecular weight of solid:

  • Q : Value of molar solution Select the

    Select the right answer of the question. Molar solution contains: (a)1000g of solute (b)1000g of solvent (c)1 litre of solvent (d)1 litre of solution

    		•
  • Q : Molarity of HCl solution 20 ml of HCL

    20 ml of HCL solution needs 19.85 ml of 0.01M NaOH solution for complete neutralization. Morality of the HCL solution is:  (i) 0.0099 (ii) 0.099 (iii) 0.99 (iv) 9.9 Choose the right answer from above.

    		•
  • Q : Preparation of ammonium sulphate Select

    Select the right answer of the question. Essential quantity of ammonium sulphate taken for preparation of 1 molar solution in 2 litres is: (a)132gm (b)264gm (c) 198gm (d) 212gm

    		•
  • Q : What do you mean by the term alum What

    What do you mean by the term alum? Also illustrate its uses?

    		•
  • Q : Concentration of urea Help me to go

    Help me to go through this problem. 6.02x 1020 molecules of urea are present in 100 ml of its solution. The concentration of urea solution is: (a) 0.02 M (b) 0.01 M (c) 0.001 M (d) 0.1 M (Avogadro constant, N4= 6.02x 1023mol -1)<

    		•
  • Q : Thermodynamics 1 Lab Report I already

    I already did Materials and Methods section. I uploaded it with the instructions. Also, make sure to see Concept Questions and Thinking Ahead in the instructions that I uploaded. deadline is tomorow at 8 am here is the link to download all instructions because I couldn't attach all of t

    		•
  • Q : What is electrolytic dissociation? The

    The Debye Huckel theory shows how the potential energy of an ion in solution depends on the ionic strength of the solution.Except at infinite dilution, electrostatic interaction between ions alters the properties of the solution from those excepted from th

    		•