(∂P/∂V)S = (∂P/∂V)T
(∂P/∂V)S = [(∂P/∂V)T]Y
(∂P/∂V)S = y(∂P/∂V)T
(∂P/∂V)S = 1/y(∂P/∂V)T
C. (∂P/∂V)S = y(∂P/∂V)T
∞
-ve
0
+ve
Vapor pressure
Specific Gibbs free energy
Specific entropy
All (A), (B) and (C)
Temperature
Pressure
Composition
All (A), (B) and (C)
1.987 cal/gm mole °K
1.987 BTU/lb. mole °R
Both (A) and (B)
Neither (A) nor (B)
μ° + RT ln f
μ°+ R ln f
μ° + T ln f
μ° + R/T ln f
Minimum number of degree of freedom of a system is zero
Degree of freedom of a system containing a gaseous mixture of helium, carbon dioxide and hydrogen is 4
For a two phase system in equilibrium made up of four non-reacting chemical species, the number of degrees of freedom is 4
Enthalpy and internal energy change is zero during phase change processes like melting, vaporisation and sublimation
More than
Less than
Equal to
Data insufficient, can't be predicted
With pressure changes at constant temperature
Under reversible isothermal volume change
During heating of an ideal gas
During cooling of an ideal gas
Accomplishes only space heating in winter
Accomplishes only space cooling in summer
Accomplishes both (A) and (B)
Works on Carnot cycle
Isothermally
Isobarically
Adiabatically
None of these
∞
+ve
0
-ve
High thermal conductivity
Low freezing point
Large latent heat of vaporisation
High viscosity
Endothermic
Exothermic
Isothermal
Adiabatic
dQ = dE + dW
dQ = dE - dW
dE = dQ + dW
dW = dQ + dE
Freon-12
Ethylene
Ammonia
Carbon dioxide
More
Less
Same
More or less; depending on the system
High temperature
Low pressure
Low temperature only
Both low temperature and high pressure
Cold reservoir approaches zero
Hot reservoir approaches infinity
Either (A) or (B)
Neither (A) nor (B)
4 J
∞
0
8 J
its internal energy (U) decreases and its entropy (S) increases
U and S both decreases
U decreases but S is constant
U is constant but S decreases
Δ S1 is always < Δ SR
Δ S1 is sometimes > Δ SR
Δ S1 is always > Δ SR
Δ S1 is always = Δ SR
Vapor compression cycle using expansion valve
Air refrigeration cycle
Vapor compression cycle using expansion engine
Carnot refrigeration cycle
-94 kcal
> -94 kcal
< - 94 kcal
Zero
Zero
50%
Almost 100%
unpredictable
(∂T/∂V)S = (∂p/∂S)V
(∂T/∂P)S = (∂V/∂S)P
(∂P/∂T)V = (∂S/∂V)T
(∂V/∂T)P = -(∂S/∂P)T
Ice at the base contains impurities which lowers its melting point
Due to the high pressure at the base, its melting point reduces
The iceberg remains in a warmer condition at the base
All (A), (B) and (C)
More than
Less than
Equal to
Not related to
More
Less
Same
Data insufficient to predict
Less than
Same as
More than
Half
Shifting the equilibrium towards right
Shifting the equilibrium towards left
No change in equilibrium condition
None of these