Chemical Engineering Thermodynamics - Study Mode

[#201] A gas can be liquified by pressure alone only, when its temperature is __________ its critical temperature.
Correct Answer

(D) Less than or equal to

Explanation

Solution: Only below and at the critical point there is strong demarcation in properties of liquids and gases above the critical point we can’t recognize the difference. Hence for a gas which is under or at the critical point can be liquefied. For a gas above critical point can also undergo a pressure change but we can’t notice the difference in properties.

[#202] In case of a close thermodynamic system, there is __________ across the boundaries.
Correct Answer

(B) No mass transfer but heat transfer

Explanation

Solution: Closed system is also called as control mass system where there is no exchange of mass between system and surroundings but energy transfer can takes place. In closed system there shouldn`t be change in identity of mass (molecules) of system even.

[#203] Clausius - Clayperon equation is applicable to __________ equilibrium processes.
Correct Answer

(D) All of the above

Explanation

Solution: clausis-claperyon equation can be applied for any equilibrium system consisting of two phases the two phases can be solid and liquid during melting or solid and vapor during sublimation and the relation is given as. $$lnP = frac{{ - Delta {H_{phase,,change}}}}{{RT}} + C$$

[#204] In any spontaneous process, the __________ free energy decreases.
Correct Answer

(C) Both A & B

Explanation

Solution: For any spontaneous process $$dG leqslant 0,,,dA leqslant 0$$ And usually spontaneous process takes place when there are finite differences so, the process will be usually irreversible. Hence $$dG < 0,,& ,,dH < 0$$ So, free energy goes on decreasing and when the finite difference becomes infinitesimal difference it becomes constant.

[#205] Gases are cooled in Joule-Thomson expansion, when it is __________ inversion temperature.
Correct Answer

(A) Below

Explanation

Solution: Above the inversion temperature joule-thomson coefficient $${left( {frac{{partial T}}{{partial P}}}
ight)_H} = - ve$$ So, by throttling (decreasing the pressure) the temperature increases and hence heating. Below the inversion temperature houle-thomson coefficient $${left( {frac{{partial T}}{{partial P}}}
ight)_H} = + ve$$ So,by throttling the temperature decreases hence cooling.