Chemical Engineering Thermodynamics - Study Mode

[#176] 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
Correct Answer

(B) Clayperon-Claussius equation

Explanation

Solution: The Clausius-Clapeyron equation for the equilibrium between liquid and vapor is then $$frac{{dp}}{{dT}} = frac{L}{{left( {Tleft( {{V_v} - {V_l}}
ight)}
ight)}}$$ 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.

[#177] 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
Correct Answer

(B) Solubility increases as temperature decreases

Explanation

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.

[#178] For the gaseous phase chemical reaction, C 2 H 4 (g) + H 2 O(g) ⟷ C 2 H 5 OH(g), the equilibrium conversion does not depend on the
Correct Answer

(D) None of these

Explanation

Solution: The equilibrium conversion is a function of temperature, pressure and concentration ratio.

[#179] For an irreversible process involving only pressure-volume work
Correct Answer

(A) (dF) T, P < 0

Explanation

Solution: Since for an spontaneous process $$Tds > dU + PdV$$ For, constant temperature and constant pressure process the above equation can be written as $$eqalign{
& dleft( {TS - U}
ight) > PdV cr
& Rightarrow dleft( { - A}
ight) > PdV cr
& Rightarrow {left( {dF}
ight)_{T,,P}} < 0. cr} $$ For spontaneous process. Here the F is Gibbs free energy and A is Helmholtz free energy.

[#180] __________ decreases during adiabatic throttling of a perfect gas.
Correct Answer

(D) Pressure

Explanation

Solution: During Throttling under adiabatic conditions the pressure energy decreases because some of the pressure energy gets lost as frictional losses thereby increase in intermolecular energy. Hence always there will be loss in pressure energy there by reducing the pressure.