--%>
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 : Q what is the basicity of primary

    what is the basicity of primary secondary and tertiary amines in chlorobenzene

    		•
  • Q : The Liver Is Responsible For Much Of

    The Liver Is Responsible For Much Of The Pentose Phosphate Activity Explain

    		•
  • Q : Linde liquefaction process Liquefied

    Liquefied natural gas (LNG) is produced using a Linde liquefaction process from pure methane gas at 3 bar and 280 K (conditions at point 1 in figure below). A three-stage compressor with interceding is used to compress the methane to 100 bar (point 2). The first stage

    		•
  • Q : Which is polar HCl or HF Which one is

    Which one is polar HCl or HF?

    		•
  • Q : Describe Thermodynamics Properties The

    The free energy property leads to convenient expressions for the volume and pressure dependence of internal energy, enthalpy and the heat capacities.All the properties of a chemical system, a sample of a substance, or a mixture of substances have some fixe

    		•
  • Q : What are biodegradable polymers?

      These are polymers that can be broken into small segments by enzyme-catalysed reactions. The required enzymes are produced by microorganism. It is a known fact that the carbon-carbon bonds of chain growth polymers are inert to enzyme-catalysed reactions, and hence they are non biod

    		•
  • Q : Solutions The relative lowering of

    The relative lowering of vapour pressure of 0.2 molal solution in which solvent is benzene

    		•
  • Q : Define thermal energy The thermal part

    The thermal part of the internal energy and the enthalpy of an ideal gas can be given a molecular level explanation. All the earlier development of internal energy and enthalpy has been "thermodynamic". We have made no use o

    		•
  • Q : Molarity of acid solution If 20ml of

    If 20ml of 0.4N, NaoH solution completely neutralises 40ml of a dibasic acid. The molarity of the acid solution is: (a) 0.1M (b) 0.2M  (c) 0.3M (d) 0.4M Choose the right answer fron above.

    		•
  • Q : Application of colligative properties

    Choose the right answer from following. Colligative properties are used for the determination of: (a) Molar Mass (b) Equivalent weight (c) Arrangement of molecules (d) Melting point and boiling point (d) Both (a) and (b)  

    		•