--%>

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 : Vapour pressure Vapour pressure of

    Vapour pressure of methanol in water Give me answer of this question. An aqueous solution of methanol in water has vapour pressure: (a) Equal to that of water (b) Equal to that of methanol (c) More than that of water (d) Less than that of water

  • Q : Mole fraction of Carbon dioxide Choose

    Choose the right answer from following. If we take 44g of CO2 and 14g of N2 what will be mole fraction of CO2 in the mixture: (a) 1/5 (b) 1/3 (c) 2/3 (d) 1/4

  • Q : Molality of Sulfuric acid Choose the

    Choose the right answer from following. The molality of 90% H2SO4 solution is: [density=1.8 gm/ml]  (a)1.8 (b) 48.4 (c) 9.18 (d) 94.6

  • Q : Facts on evaporation Illustrate the 3

    Illustrate the 3 facts on evaporation?

  • Q : 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

  • Q : Sugar solution The solution of sugar in

    The solution of sugar in water comprises: (i) Free atoms (ii) Free ions (iii) Free molecules (iv) Free atom and molecules. Choose the right answer from the above.

  • Q : Problem on decinormal strength Can

    Can someone please help me in getting through this problem. How many grams of dibasic acid (having mol. wt. 200) must be present in 100ml  of its aqueous solution to provide decinormal strength: (i) 1g  (ii)2g  (iii) 10g  (iv) 20g<

  • Q : What is Flash Photolysis Reactions.

    An example illustrates the type of mechanism that can be written to explain the development of flash photolysis reactions. Often, as the reactions in the ozone layer of the earth's atmosphere, we are interested in the kinetic behavior of species that are not a

  • Q : Benefits of soapy detergents over the

    What are the benefits of soapy detergents over the soap less detergents? Briefly state the benefits?

  • Q : Group IV Cations Chromium(III)

    Chromium(III) hydroxide is highly insoluble in distilled water but dissolves readily in either acidic or basic solution. Briefly explain why the compound can dissolve in acidic or in basic but not in neutral solution. Write appropriate equations to support your answer.