Chemical Engineering Thermodynamics MCQ Questions & Answers | Chemical Engineering

Definition of heat capacity: the amount of heat required to raise the temperature of a substance or material by a small amount is called heat capacity.

$$\mathop {\lim }\limits_{dT \to 0} \,dQ = C\,\left( {{\text{heat capacity}}} \right)$$

Every substance has its own molecules and every molecule in a substance has its own properties and structure depending on the nature of the substance so different molecules exhibit different heat capacity at the same temperature. But every substance molecules at 0K will come to rest and to raise the temperature of a substance by a negligible amount doesn’t depend upon the nature of the substance but once the substance reaches $$dT$$ amount of temperature different substance or materials exhibit different heat capacity.

We know, the property relation:
$$dG = vdp - sdT$$
Which is valid for both reversible and irreversible process since it is a property relation. when the system undergoes isothermal change $$dT = 0$$
$$\eqalign{ & {\text{So, }}dG = vd \cr & \Rightarrow dG = \frac{{nRT}}{P}dP\left( {{\text{for, ideal gas}}} \right) \cr} $$
On, integration
$$ \Rightarrow \Delta G = nRT.ln\frac{{{P_2}}}{{{P_1}}}$$
So, for a system containing ideal gas and undergoing isothermal change of volume or pressure this expression is valid.

Since $$P{V^Y} = $$   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 $$P{V^Y} = $$   constant can’t be valid.

$$A.$$ solid ⟹ liquid → melting
$$B.$$ liquid ⟹ vapor → vaporization
$$C.$$ solid ⟹ vapor → sublimation
So, enthalpy change for the reaction $${H_2}O\left( {liq} \right) \Rightarrow {H_2}O\left( {gas} \right)$$     is enthalpy change of vaporization.
In the question entropy should be replaced by enthalpy.

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.

Activity coefficient measures the extent to which a real gas deviates from ideality.
Hence, for ideal gas activity coefficient = 1.

The internal energy ($$U$$) is a function of

$$dU = CvdT - \left[ {P + T\left\{ {\frac{{\left( {\frac{{\partial V}}{{\partial T}}} \right)p}}{{\left( {\frac{{\partial V}}{{\partial P}}} \right)T}}} \right\}dV} \right]$$

For an ideal gas, $$PV = RT$$

So, $$\left( {\frac{{\partial V}}{{\partial T}}} \right)p = \frac{R}{P}{\text{ and}}\left( {\frac{{\partial V}}{{\partial T}}} \right)T = \frac{{ - RT}}{{{P^2}}}$$

Hence, $$dU = CvdT$$
So, for an ideal gas if it undergoing isothermal change $$\left( {dT = 0} \right) \Rightarrow dU = 0$$
So, the questioned should be changed and should be mentioned for an ideal gas.

According to law of corresponding states if two gases have same reduced temperature and reduced pressure than they will have same reduced volume.
Law of corresponding state: $$\left[ {Pr + \frac{3}{{{v_r}^2}}} \right]\left[ {3Vr - 1} \right] = 8R{T_r}$$
So, the answer is reduced volume.

Gibbs phase rule is, $$G = C - \in + 2$$
Where :
$$C$$ = Number of components
$$ \in $$ = Number of phases
$$G$$ = Degree of freedom
Clearly we can see Gibbs phase rule is more useful when two or more phases are involved in the given options the condensation is the only process where more than one phase is involved so, the Gibbs phase rule finds its application there.

The phase change occurs at constant temperature and pressure under normal situations. But the volume changes because between two phases always there is a change in intermolecular forces except at critical condition.
Eg: when liquid is converting to vapor the vapor formed will have more volume than liquid.