--%>
Assignment Help - homework help

Problem on reversible process

a. For a reversible process involving ideal gases in a closed system, Illustrate that

ΔS = Cv ln(T2/T1) for a constant volume process
ΔS = Cp ln(T2/T1) for a constant pressure process
ΔS = R ln(V2/V1) = R ln(P1/P2) for an isothermal process

b. Determine the change in entropy when 2 kg of a gas at 277 K is heated in a closed vessel to a temperature of 368 K. Taking specific heat at constant volume to be 1.42 KJ/Kg.K

c. Compute the entropy change when 1 kmol of an ideal gas at 300 K and 10 bar expands to a pressure of 1 bar if the temperature remains constant.

d. What is the change in entropy when 1 kmol of an  ideal gas at 335 K and 10 bar is expanded irreversibly to 300 K and 1 bar? Cp = 29.3 KJ/(kmol K)

   Related Questions in Chemistry

  • Q : Particles of quartz Particles of quartz

    Particles of quartz are packed by:(i) Electrical attraction forces  (ii) Vander Waal's forces  (iii) Covalent bond forces  (iv) Strong electrostatic force of attraction Answer: (iii)

    		•
  • Q : Relative lowering of vapour pressure

    explain the process of relative lowering of vapour pressure

    		•
  • Q : C-X bond length in halobenzene less

    C-X bond length in halobenzene less then C-X bond lengthin CH3-x

    		•
  • Q : How haloalkanes are prepared from

    Alkyl halides can be prepared from alkanes through substitution and from alkenes through addition of halogen acids or through allylic substitution.    From alkanesWhen alkanes are treated with halogens, chlo

    		•
  • Q : Wavelengths which the human eye can see

    Briefly state the wavelengths which the human eye can see?

    		•
  • Q : Analytical chemistry 37% weight of HCl

    37% weight of HCl and density is 1.1g/ml. find molarity of HCl

    		•
  • Q : Problem on Osmotic Pressure of solution

    The osmotic pressure of a 5% solution of cane sugar at 150oC  is (mol. wt. of cane sugar = 342)(a) 4 atm (b) 3.4 atm (c) 5.07 atm (d) 2.45 atmAnswer: (c) Π = (5 x 0.0821 x 1000 x 423)/(342 x 100) = 5.07 atm

    		•
  • Q : Calculation of molecular weight Provide

    Provide solution of this question. In an experiment, 1 g of a non-volatile solute was dissolved in 100 g of acetone (mol. mass = 58) at 298K. The vapour pressure of the solution was found to be 192.5 mm Hg. The molecular weight of the solute is (vapour pressure of ace

    		•
  • Q : Question based on vapour pressure While

    While a substance is dissolved in a solvent, the vapour pressure of the solvent is decreased. This results in: (a) An increase in the boiling point of the solution (b) A decrease in the boiling point of solvent (c) The solution having a higher freezing point than

    		•
  • Q : Utilization of glacial acetic acid What

    What is the utilization of glacial acetic acid? Briefly describe the uses.

    		•