Chemical Engineering Thermodynamics - Study Mode

[#371] What is the ratio of adiabatic compressibility to isothermal compressibility?
Correct Answer

(B) < 1

Explanation

Solution: By T-Ds Equations at constant entropy $$eqalign{
& {C_p}dT = Tfrac{{partial V}}{{partial {T_P}}}dP{ ext{ and}} cr
& {C_v} = - T{left( {frac{{partial P}}{{partial T}}}
ight)_P}{left( {frac{{partial V}}{{partial T}}}
ight)_S} cr
& Rightarrow frac{{{C_P}}}{{{C_V}}} = frac{{left( {frac{{partial P}}{{partial V}}}
ight)S}}{{left( {frac{{partial P}}{{partial V}}}
ight)T}} cr} $$ Since, $${C_P}$$ xa0is always greater than $${C_V}$$ xa0the ratio of isothermal compressibility and isentropic (reversible adiabatic) process is always greater than $$1 Rightarrow $$ xa0 the difference is greater than zero.

[#372] Which of the following non-flow reversible compression processes require maximum work?
Correct Answer

(A) Adiabatic process

Explanation

Solution: If we see the $$P-V$$ xa0plots for isobaric, adiabatic and isothermal process the area under the graph is more in case of isobaric process hence the work done in isobaric process is maximum.

[#373] What is the value of Joule-Thomson co-efficient for an ideal gas?
Correct Answer

(C) 0

Explanation

Solution: The joule Thomson coefficient is given as $${mu _i} = {left( {frac{{partial T}}{{partial P}}}
ight)_H},$$ xa0 xa0And since for an ideal gas enthalpy is strictly only function of temperature which implies constant temperature and hence the joule thomson coefficient becomes zero. Physically both the contervining effects of throttling are balancing each other.

[#374] Which of the following equations is used for the prediction of activity co-efficient from experiments?
Correct Answer

(D) All of the above

Explanation

Solution: All the given three equations are models to find activity co-efficient.

[#375] The molar excess Gibbs free energy, $${{ ext{G}}^{ ext{E}}}$$, for a binary liquid mixture at T and P is given by, $$left( {frac{{{{ ext{G}}^{ ext{E}}}}}{{{ ext{RT}}}}}
ight)$$ xa0= Ax 1 x 2 , where A is a constant. The corresponding equation for ln y 1 , where y 1 is the activity co-efficient of component 1, is
Correct Answer

(A) Ax 2 2

Explanation

Solution: Given, $$frac{{{G^E}}}{{RT}} = A{x_1}{x_2} Rightarrow frac{{n{G^E}}}{{RT}} = frac{{A{n_1}{n_2}}}{n}$$ On partially differentiating it $$left( {frac{{partial frac{{n{G^E}}}{{RT}}}}{{partial {n_1}}}}
ight)T,P,{n_2} = frac{{A{n_1}{n_2}}}{n} = A{x_2}^2$$ As $$left( {frac{{partial frac{{n{G^E}}}{{RT}}}}{{partial {n_1}}}}
ight)T,P,{n_2} = frac{{overline {n{G^E}_l} }}{{RT}} = ln{gamma _1}$$ Hence $$ln{gamma _1} = A{x_2}^2$$