Entropy
Gibbs energy
Internal energy
Enthalpy
Pressure remains constant
Pressure is increased
Temperature remains constant
None of these
Freon
Liquid sulphur dioxide
Methyl chloride
Ammonia
The available energy in an isolated system for all irreversible (real) processes decreases
The efficiency of a Carnot engine increases, if the sink temperature is decreased
The reversible work for compression in non-flow process under isothermal condition is the change in Helmholtz free energy
All (A), (B) and (C)
Volume
Temperature
Pressure
None of these
Less than
More than
Same as
Not related to
Turbine
Heat engine
Reversed heat engine
None of these
3
1
2
0
Decrease in velocity
Decrease in temperature
Decrease in kinetic energy
Energy spent in doing work
Unity
Activity
Both (A) & (B)
Neither (A) nor (B)
First law
Zeroth law
Third law
Second law
Is the most efficient of all refrigeration cycles
Has very low efficiency
Requires relatively large quantities of air to achieve a significant amount of refrigeration
Both (B) and (C)
Latent heat of vaporisation
Chemical potential
Molal boiling point
Heat capacity
With pressure changes at constant temperature
Under reversible isothermal volume change
During heating of an ideal gas
During cooling of an ideal gas
Molecular size
Volume
Pressure
Temperature
A closed system does not permit exchange of mass with its surroundings but may permit exchange of energy.
An open system permits exchange of both mass and energy with its surroundings
The term microstate is used to characterise an individual, whereas macro-state is used to designate a group of micro-states with common characteristics
None of the above
Increase the partial pressure of I2
Decrease the partial pressure of HI
Diminish the degree of dissociation of HI
None of these
Less
More
Same
Dependent on climatic conditions
Low pressure & high temperature
High pressure & low temperature
Low pressure & low temperature
None of these
Specific heat
Latent heat of vaporisation
Viscosity
Specific vapor volume
ΔF = ΔH + T [∂(ΔF)/∂T]P
ΔF = ΔH - TΔT
d(E - TS) T, V < 0
dP/dT = ΔHvap/T.ΔVvap
Low pressure and high temperature
Low pressure and low temperature
High pressure and low temperature
High pressure and high temperature
Specific volume
Work
Pressure
Temperature
Process must be isobaric
Temperature must decrease
Process must be adiabatic
Both (B) and (C)
0
2
1
3
d ln p/dt = Hvap/RT2
d ln p/dt = RT2/Hvap
dp/dt = RT2/Hvap
dp/dt = Hvap/RT2
Melting point of ice
Melting point of wax
Boiling point of liquids
None of these
Mass
Momentum
Energy
None of these
Molecular size
Temperature
Volume
Pressure
0
1
2
3