Download Tutorial II (thermodynamics)

Tutorial II(thermodynamics)
(CYL110/CY110N)
Q1. Read the following statements carefully and decide whether (and why) they are true
or false giving your thermodynamic reasoning. If the statement is false you may either
state which law or laws of thermodynamics it violates or provide a physical counter
example or any other plausible physical reason. Finally, correct the false statement with a
clarifying phrase that makes the statement true.
(a) No heat transfer occurs when liquid water is reversibly and isothermally compressed.
(b) During an adiabatic, reversible process at constant volume the internal energy can
never increase.
(c) The internal energy of a system and its surroundings is never conserved during an
irreversible process, but is conserved for reversible processes.
(d) The work done by a closed system can exceed the decrease in the system’s internal
energy
Q2. An ideal monoatomic gas (Cv,m= 1.5R) initially at 298 K and 1.013 Mpa pressure
expands adiabatically and irreversibly until it is in equilibrium with constant external
pressure of 0.1013 Mpa. What is the final temperature of the gas?
Q3. One mole of chlorine undergoes adiabatic expansion from 1 dm3 to 10 dm3 against a
constant external pressure of 1 bar. The initial temperature of the gas was 273 K.
Calculate q, w, ∆U and T2 if the gas is considered a van der waals gas.
Given a= 655 dm6 Kpa mol-2
B=0.055 dm3 mol-1
Cv,m= 33.91 JK-1mol-2.
Q4: One mole of a gas is expanded isothermally at 298 K until its volume is tripled. Find
the value of ∆Sgas and ∆Stotal under the following conditions?
(a)Expansion is carried out reversibly.
(b) Expansion is carried out irreversibly where 836.8 J of heat is less absorbed than in
case (a).
© Expansion is free.
Q5: A sample of 1.00 mole of monoatomic perfect gas with Cv,m = 1.5 R, initially at 298
K and 10 L, is expanded, with the surroundings maintained at 298 K, to a final volume of
20 L, in three ways (a) isothermally and reversibly (b) isothermally against a constant
external pressure of 0.5 atm (c) adiabatically against a constant external pressure of 0.5
atm . Calculate ∆S and ∆Ssurr, ∆H and ∆T for every path.
Q6. A 50 gm mass of Cu at a temperature of 393 K is placed in contact with a 100 g mass
of Cu at a temperature of 303 in a thermally insulated container. Calculate q and ∆Stotal
for the reversible process. Use a value of 0.4184 Jg-1K-1 for specific heat capacity of Cu.
Q7. Calculate the entropy change at 373 K for the transformation.
H2O (l, 1.01325 bar) = H2O (g, 0.101325)
Given ∆Hvap = 40.668 kJmol-1
Q8. 5 gm of ice at 273 K is added to the 30 gm of water at 323 K in a thermally insulated
container (a) what is the final temperature? (b) What is the total entropy change?
∆Hfus of ice at 273 K = 334.72 J/gm
∆Hvap of water at 373 K= 2259.26 J/gm
Average heat capacity of the liquid water = 4.184 JK-1gm-1
Q9. Calculate ∆G for the conversion of 3 mol of liquid benzene at 80 °C (normal boiling
point) to vapour at same temperature and a pressure of 0.66 bar. Consider the vapour as
an ideal gas.
Q10. In the transition CaCO3 (aragonite) to CaCO3 (calcite), ∆Gm (298) = -800 J and
∆Vm = 2.75 cm3. At what pressure would aragonite become the stable form at 298 K.
Q11. Calculate the change in the molar Gibbs energy of (a) liquid water treated as an
incompressible fluid and (b) the water vapour treated as a perfect gas, when the pressure
is increased isothermally from 1 bar to 2 bar at 298 K.
Q12. For H2O (l) = H2O (g)
∆G298K=8.596 kJmol-1
∆H = 57086.5 – 45.327 (T) +4.966 * 10-3 (T)2
Compute ∆G298K.
Q13. Calculate the effect on the chemical potentials of ice and water of increasing the
pressure from 1 bar to 2 bar at 0 °C. The density of ice is 0.917 gcm-3 and that of liquid
water is 0.999 gm cm-3 under these conditions.
Q14. What is the change in the boiling point of water at 100 °C per Pa change in
atmospheric pressure? Molar enthalpy of vaporization is 40.69 kJ mol-1?
Q15. Calculate the equilibrium pressure for the conversion of graphite to diamond at 25
C. The densities of graphite and diamond may be taken to be 2.25 and 3.51 g cm-3,
respectively, independent of pressure, in calculating the change of ∆G with pressure. ∆G°
= 2900 J mol-1.