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1
Pick out the wrong statement.
Discuss
Answer & Solution
Answer: Option C
Solution:
Since the chemical potential of the specie in a mixture is given by $${\mu _i} = \mu _i^0 + RT\,\ln \left( {\frac{{{P_i}}}{{1\,bar}}} \right)$$     hence when pressure tends to zero chemical potential of the specie tends to infinite.
2
As pressure approaches zero, the ratio of fugacity to pressure $$\left( {\frac{{\text{f}}}{{\text{P}}}} \right)$$  for a gas approaches
Discuss
Answer & Solution
Answer: Option B
Solution:
As pressure approaches zero since gas approaches ideal behavior the ratio of fugacity and pressure becomes unity.
3
The theoretical minimum work required to separate one mole of a liquid mixture at 1 atm, containing 50 mole % each of n- heptane and n- octane into pure compounds each at 1 atm is
Discuss
Answer & Solution
Answer: Option B
Solution:
Entropy of mixing is
$$\eqalign{ & = - R\sum {{X_i}\ln {x_i}} \cr & = - R\left[ {0.5\,\ln \,0.5 + 0.5\,\ln \,0.5} \right] \cr} $$
Since, work $$ = T\Delta S = - RT\,\ln \,0.5.$$
4
When liquid and vapour phases of one component system are in equilibrium (at a given temperature and pressure), the molar free energy is
Discuss
Answer & Solution
Answer: Option C
Solution:
If the liquid and vapor are pure, in that they consist of only one molecular component and no impurities, then the equilibrium state between the two phases is described by the following equations:
Pliq = pvap, TLIQ = TVAP, GLIQ = GVAP.
5
If the molar heat capacities (Cp or Cv) of the reactants and products of a chemical reaction are identical, then, with the increase in temperature, the heat of reaction will
Discuss
Answer & Solution
Answer: Option C
Solution:
The heat of reaction will remain unaltered when molar heat capacities of reactants and products of a chemical reaction remain same.
6
If the vapour pressure at two temperatures of a solid phase in equilibrium with its liquid phase are known, then the latent heat of fusion can be calculated by the
Discuss
Answer & Solution
Answer: Option B
Solution:
The Clausius-Clapeyron equation for the equilibrium between liquid and vapor is then
$$\frac{{dp}}{{dT}} = \frac{L}{{\left( {T\left( {{V_v} - {V_l}} \right)} \right)}}$$
Where $$L$$ is the latent heat of evaporation, and $${{V_v}}$$ and $${{V_l}}$$ are the specific volumes at temperature $$T$$ of the vapor and liquid phases, respectively.
More generally the Clausius-Clapeyron equation pertains to the relationship between the pressure and temperature for conditions of equilibrium between two phases. The two phases could be vapor and solid for sublimation or solid and liquid for melting.
7
If the heat of solution of an ideal gas in a liquid is negative, then its solubility at a given partial pressure varies with the temperature as
Discuss
Answer & Solution
Answer: Option B
Solution:
The effect of temperature on the solubility of a gas in a liquid:
According to Charles's law, volume of a given mass of a gas increases with increase in temperature. The volume of given mass of dissolved gas in solution also increases with increase of temperature. It becomes impossible for solvent to accommodate gaseous solute in it and gas bubbles out. Hence, with increase in temperature, the solubility of a gas in a liquid decreases.
8
Pick out the wrong statement.
Discuss
Answer & Solution
Answer: Option C
Solution:
The equation relating pressure, temperature and volume is called as equation of state and ideal gas equation is one of the examples of equation of state.
9
For the gaseous phase chemical reaction, C2H4(g) + H2O(g) ⟷ C2H5OH(g), the equilibrium conversion does not depend on the
Discuss
Answer & Solution
Answer: Option D
Solution:
The equilibrium conversion is a function of temperature, pressure and concentration ratio.
10
For an irreversible process involving only pressure-volume work
Discuss
Answer & Solution
Answer: Option A
Solution:
Since for an spontaneous process $$Tds > dU + PdV$$
For, constant temperature and constant pressure process the above equation can be written as
$$\eqalign{ & d\left( {TS - U} \right) > PdV \cr & \Rightarrow d\left( { - A} \right) > PdV \cr & \Rightarrow {\left( {dF} \right)_{T,\,P}} < 0. \cr} $$
For spontaneous process.
Here the F is Gibbs free energy and A is Helmholtz free energy.