Chemical Engineering Thermodynamics - Study Mode
[#101] The thermodynamic law, PV Y = constant, is not applicable in case of
Correct Answer
(B) Free expansion of an ideal gas
Explanation
Solution: Since $$P{V^Y} = $$ xa0 constant is valid only for reversible process but since as free expansion is irreversible because it is working on a cycle by taking heat from a single reservoir and producing net expansion work when considered reversible and thus violating Kelvin-planck statement when considered reversible hence we can conclude free expansion is an irreversible process. And hence $$P{V^Y} = $$ xa0 constant can’t be valid.
[#102] Entropy change of the reaction, H 2 O(liquid) ⇒ H 2 O(gas), is termed as the enthalpy of
Correct Answer
(B) Vaporisation
Explanation
Solution: $$A.$$ solid ⟹ liquid → melting $$B.$$ liquid ⟹ vapor → vaporization $$C.$$ solid ⟹ vapor → sublimation So, enthalpy change for the reaction $${H_2}Oleft( {liq}
ight) Rightarrow {H_2}Oleft( {gas}
ight)$$ xa0 xa0 is enthalpy change of vaporization. In the question entropy should be replaced by enthalpy.
[#103] Specific __________ does not change during a phase change (e.g. sublimation, melting, vaporisation etc.).
Correct Answer
(D) Gibbs free energy
Explanation
Solution: During phase change suppose consider liquid to vapor the entropy increases because randomness increases and the liquid converts to vapor by taking enthalpy so, enthalpy changes and generally the enthalpy of vapor will be greater than liquid. We may think internal energy remains constant during phase change since temperature remains constant but here the potential energy changes thereby, changing internal energy since in vapor the distance between molecules is greater than the distance between the molecules in liquid hence the work made is different. So internal energy is different. But we know Gibbs free energy is a function of pressure and temperature and during the phase change since pressure and temperature remains constant so, Gibbs free energy remains constant, but not zero.
[#104] For an ideal gas, the activity co-efficient is
Correct Answer
(C) Unity at all pressures
Explanation
Solution: Activity coefficient measures the extent to which a real gas deviates from ideality. Hence, for ideal gas activity coefficient = 1.
[#105] For an isothermal process, the internal energy of a gas
Correct Answer
(C) Remains unchanged
Explanation
Solution: The internal energy ($$U$$) is a function of $$dU = CvdT - left[ {P + Tleft{ {frac{{left( {frac{{partial V}}{{partial T}}}
ight)p}}{{left( {frac{{partial V}}{{partial P}}}
ight)T}}}
ight}dV}
ight]$$ For an ideal gas, $$PV = RT$$ So, $$left( {frac{{partial V}}{{partial T}}}
ight)p = frac{R}{P}{ ext{ and}}left( {frac{{partial V}}{{partial T}}}
ight)T = frac{{ - RT}}{{{P^2}}}$$ Hence, $$dU = CvdT$$ So, for an ideal gas if it undergoing isothermal change $$left( {dT = 0}
ight) Rightarrow dU = 0$$ So, the questioned should be changed and should be mentioned for an ideal gas.