Chemical Engineering Thermodynamics - Study Mode
[#406] The expression for entropy change given by, $$Delta { ext{S}} = { ext{nR}}l{ ext{n}}left( {frac{{{{ ext{V}}_2}}}{{{{ ext{V}}_1}}}}
ight) + { ext{n}}{{ ext{C}}_{ ext{v}}}l{ ext{n}}left( {frac{{{{ ext{T}}_2}}}{{{{ ext{T}}_1}}}}
ight)$$ xa0 xa0 xa0is valid for
Correct Answer
(D) Simultaneous heating and expansion of an ideal gas
Explanation
Solution: For a pure substance in a closed system has a degree of freedom = 2 So, entropy(s) can be expressed as a function of two variables (properties) $$S = fn(T,V)$$ [x08egin{align}
& Rightarrow dS={{left( frac{partial S}{partial T}
ight)}_{V}}dT+{{left( frac{partial S}{partial v}
ight)}_{T}}dT \
& Rightarrow dS=frac{{{C}_{V}}dT}{T}+{{left( frac{partial P}{partial T}
ight)}_{V}}dVleft( ext{for one mole of ideal gas}{{left( frac{partial P}{partial T}
ight)}_{V}}=frac{R}{V}
ight) \
& Rightarrow dS=frac{{{C}_{V}}}{T}dT+frac{nR}{V}dV \
end{align}] So, on integrating this equation from temperature T 2 to T 1 expanding volume from V 1 to V 2 GIVES the equation $$Delta S = nRlnleft( {frac{{{V_2}}}{{{V_1}}}}
ight), + nCvlnleft( {frac{{{T_2}}}{{{T_1}}}}
ight)$$ for an ideal gas of n moles.
[#407] On a P-V diagram of an ideal gas, suppose a reversible adiabatic line intersects a reversible isothermal line at point A. Then at a point A, the slope of the reversible adiabatic line $${left( {frac{{partial { ext{P}}}}{{partial { ext{V}}}}}
ight)_{ ext{S}}}$$ xa0and the slope of the reversible isothermal line $${left( {frac{{partial { ext{P}}}}{{partial { ext{V}}}}}
ight)_{ ext{T}}}$$ xa0are related as (where, $${ ext{y}} = frac{{{{ ext{C}}_{ ext{p}}}}}{{{{ ext{C}}_{ ext{v}}}}}$$ xa0)
Correct Answer
(C) $${left( {frac{{partial { ext{P}}}}{{partial { ext{V}}}}}
ight)_{ ext{S}}} = { ext{y}}{left( {frac{{partial { ext{P}}}}{{partial { ext{V}}}}}
ight)_{ ext{T}}}$$
Explanation
Solution: For an adiabatic process → $$P{V^y}$$ = constant For an isothermal process → $$PV$$ = constant So, slope for adiabatic process in $$PV$$ plane is $$frac{{dp}}{{dv}} = - yfrac{p}{v}$$ Slope for isothermal process is $$frac{{dp}}{{dv}} = - frac{p}{v}$$ Hence, $${left( {frac{{partial p}}{{partial v}}}
ight)_S} = - y{left( {frac{{partial p}}{{partial v}}}
ight)_T}$$
[#408] Trouton's ratio is given by (where $${lambda _{ ext{b}}}$$ = molal heat of vaporisation of a substance at its normal boiling point, kcal/kmol T b = normal boiling point, °K )
Correct Answer
(A) $$frac{{{lambda _{ ext{b}}}}}{{{{ ext{T}}_{ ext{b}}}}}$$
Explanation
Solution: Troutons rule states that for most of the liquid the entropy of vaporisation is constant at boiling points and the entropy of vaporisation is given by ratio of enthalpy of vaporisation to normal boiling temperature $$left( {{T_b}}
ight)$$ nothing but troutons ratio. $$Delta {S_{vap}} = frac{{Delta {H_{vap}}}}{{{T_b}}}$$
[#409] Isentropic process means a constant __________ process.
Correct Answer
(C) Entropy
Explanation
Solution: Isentropic process is the process where entropy(s) remains constant. So, $$ds = 0$$ Generally, $$ds$$ $$ = frac{{delta Q}}{T}$$ xa0 for reversible process, so for reversible adiabatic process $$ds = 0$$ Hence, reversible adiabatic process is called isentropic process.
[#410] Sublimation temperature of dry ice (solid CO 2 ) is __________ °C.
Correct Answer
(C) -78
Explanation
Solution: Sublimation means the conversion of solid phase directly in to vapor phase (without involvement of phase liquid). Actually for every substance this direct phase conversion occurs when the substance is below its triple point pressure and corresponding temperature. Actually one of the point where the dry ice sublimates is 1 atm pressure and corresponding temperature is $$ - {78^ circ }C.$$