--%>
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 : 6. 20 gm of hydrogen is present in 5

    6. 20 gm of hydrogen is present in 5 litre vessel. The molar concentration of hydrogen is

    		•
  • Q : Describe physical adsorption and its

    When the forces of attraction existing between adsorbate and adsorbent are van der Waal's forces, the adsorption is called physical adsorption. This type of adsorption is also known as physisorption or van der Waal's adsorption. Since the forces existing between adsorbent and adsorbate are very w

    		•
  • Q : Dipole attractions for london dispersion

    Illustrate how are dipole attractions London dispersion forces and hydrogen bonding similar?

    		•
  • Q : Basicity order order of decreasing

    order of decreasing basicity of urea and its substituents

    		•
  • Q : Molar concentration of Iron chloride

    Provide solution of this question. A certain aqueous solution of FeCl3 (formula mass =162) has a density of 1.1g/ml and contains 20.0% Fecl. Molar concentration of this solution is: (a) .028 (b) 0.163 (c) 1.27 (d) 1.47

    		•
  • Q : Molar mass Select the right answer of

    Select the right answer of the question. Which is heaviest: (a)25 gm of mercury (b)2 moles of water (c)2 moles of carbon dioxide (d)4 gm atoms of oxygen

    		•
  • Q : Problem on partial pressure i) Show

    i) Show that the equilibrium constant Kp for the reaction CaCo3(s) ↔ CaO(s) +CO2(g)is about unity (i.e. = 1.0) at 895 °C.ii) If two grams of calcium carbonate are pl

    		•
  • Q : Symmetry Elements The symmetry of the

    The symmetry of the molecules can be described in terms of electrons of symmetry and the corresponding symmetry operations.Clearly some molecules, like H2O and CH4, are symmetric. Now w

    		•
  • Q : Isotonic Solutions Which one of the

    Which one of the following pairs of solutions can we expect to be isotonic at the same temperature:(i) 0.1M Urea and 0.1M Nacl  (ii) 0.1M Urea and 0.2M Mgcl2  (iii) 0.1M Nacl and 0.1M Na2SO4  (iv) 0.1M Ca(NO3<

    		•
  • Q : Ddd 4) The addition of S2- ion to

    4) The addition of S2- ion to Fe(OH)2(s). Explain why the addition of S2- ion to Cr(OH)3(s) does not result in the formation of Cr2S3(s).

    		•