Properties of Pure Substances MCQ Quiz - Objective Question with Answer for Properties of Pure Substances - Download Free PDF

Concept:

Vacuum efficiency is given by:

$\text{Vacuum Efficiency}=\frac{\text{Actual Vacuum}}{\text{Ideal Vacuum}}\times 100$

Given:

• Barometer reading = 760 mm Hg
• Vacuum recorded in condenser = 690 mm Hg
• Saturation pressure at 35°C = 42.21 mm Hg

Calculation:

Ideal Vacuum = $760-42.21=717.79\text{mm Hg}$

Vacuum Efficiency = $\frac{690}{717.79}\times 100=96.10\mathrm{\%}$

Explanation:

Mollier diagram:

• Mollier chart is drawn between enthalpy (h) and entropy (s).
• The turbine output on the Mollier diagram represented by the vertical straight line in ideal conditions.
• On Mollier diagram constant pressure line diverge from one another as shown in the figure.

From the first and second laws of thermodynamics, the following property relation was obtained,

Tds = dh – vdp

${\left(\frac{\partial h}{\partial s}\right)}_p=T$

This equation forms the basis of the h-s diagram of a pure substance, also called the Mollier diagram. The slope of an isobar on the h-s coordinates is equal to the absolute saturation temperature ( tsat + 273) at that pressure.

Mollier diagram is enthalpy (h) versus entropy (s) plot.

It consists of a family of constant pressure lines, constant temperature lines and constant volume lines plotted on enthalpy versus entropy coordinates.

The slope of the constant pressure line on an h-s curve is equal to absolute temperature and due to this reason, the constant pressure lines are of diverging nature in the superheated region.

Concept:

The equation that relate partial derivatives of properties of p, v, T and s of a compressible fluid are called Maxwell relations.

The four Gibbsian relations for a unit mass are

1) du = Tds – Pdv

2) dh = Tds + vdP

3) df = - Pdv – sdT

4) dg = -sdT + vdP

Since u,h,f and g are the properties thus point functions and the above relations can be expressed as

dz = Mdx + Ndy

with,

${\left(\frac{\partial T}{\partial v}\right)}_s=+{\left(\frac{\partial p}{\partial s}\right)}_v$

Applying the cyclic relation as

Mdx + Ndy → ${\left(\frac{\delta M}{\delta y}\right)}_x={\left(\frac{\delta N}{\delta x}\right)}_y$

Now,

Replacing M,N,y and x by T,p,v,s of each of the Gibbsian equations in cyclic order, we will get the following four relations.

$1){\left(\frac{\partial T}{\partial p}\right)}_s={\left(\frac{\partial v}{\partial s}\right)}_p$

$2){\left(\frac{\partial p}{\partial T}\right)}_v={\left(\frac{\partial s}{\partial v}\right)}_T$

$3){\left(\frac{\partial T}{\partial v}\right)}_s=-{\left(\frac{\partial p}{\partial s}\right)}_v$

$4){\left(\frac{\partial v}{\partial T}\right)}_p=-{\left(\frac{\partial s}{\partial p}\right)}_T$

Concept:

Combined equations of the first and second law of thermodynamics

Tds = du + Pdv

Tds = dh – vdP

These equations are applicable for both reversible and irreversible process and for the closed and open system as well.

du = CvdT

dh = CpdT

Cv = specific heat at constant volume, Cp = specific heat at constant pressure

From second equation

Tds = dh – vdP

For constant pressure, dP = 0 & dh = CpdT

Tds = CpdT

${\frac{dT}{dS}|}_{p=c}=\frac{T}{C_p}$

Hence on the T-S diagram, the slope of the constant pressure line = T/Cp

Additional Information

From first equation

Tds = du + Pdv

For constant volume, dv = 0

And, du = CvdT

∴ 1st equation becomes: Tds = CvdT

${\frac{dT}{dS}|}_{v=c}=\frac{T}{C_v}$

Concept:

Total volume of the tank is the sum of volume of liquid water and volume of saturated vapor.

Since the specific volume of liquid water is very small (~0.001 m³/kg), its volume is negligible.

Calculation:

$V_{\text{liquid}}=100\times 0.001=0.1{\text{m}}^3$

$V_{\text{vapor}}=5\times 57.8=289{\text{m}}^3$

$V_{\text{total}}=0.1+289=289.1\approx 290{\text{m}}^3$

$v=v_f+x\left(v_g-v_f\right)$

v = V/m = 0.5 m³/kg

$0.5=0.00106+x\left(0.8908-0.00106\right)$

$x=0.56$

$0\le x\le 1$

x = 0, Saturated liquid

x = 1, Saturated vapor

As x value lies between 0 and 1, therefore it is a mixture of saturated liquid and saturated vapor.

Explanation:

Since, the properties like internal energy, enthalpy and entropy of a system cannot be directly measured. They are related to change in the energy of the system.

Hence, we can determine Δu, Δh, Δs but not the absolute values of these properties.

⇒ Therefore, It is necessary to choose a reference state to which, these properties are arbitrary assigned some numerical values.  

So for water, the triple point (T = 0.01°C & P = 611 Pa) is selected as reference a state, where the “Internal energy” (u) and “Entropy” (s) of saturated liquid are assigned a zero value.

#Note: h = u + Pv

At triple point, u = 0, but p × ν ≠ 0

Explanation:

Joule – Thomson coefficient:- When the gas in steady flow passes through a constriction, e.g. in an orifice or valve, it normally experiences a change in temperature. From the first law of thermodynamics, such a process is isenthalpic and one can usefully define a Joule – Thomson coefficient as

$\mu ={\left(\frac{\partial T}{\partial P}\right)}_H$

As a measure of the change in temperature which results from a drop in pressure across the construction.

• For an ideal gas, μ = 0, because ideal gases neither warm not cool upon being expanded at constant enthalpy.
• If μ is +ve, then the temperature will fall during throttling.
• If μ is -ve, then the temperature will rise during throttling.

Explanation

The phase diagram of water is given below.

A substance sublimates on heating If it is kept below its triple point pressure. Hence the lowest pressure at which water can exist in liquid phase in stable equilibrium is triple point pressure.

The triple point properties of water are given below.

Triple point pressure, P­_tp = 4.58 mm of Hg = 0.611 kPa

Triple point temperature, T_tp = 273.16 K = 0.01°C

Concept:

Enthalpy is sum of internal energy plus pV work.

h = u + pv

Calculation:

Given data: p = 8 MPa = 8 × 103 kPa, u = 2864 kJ/kg, v = 0.03432 m3/kg

h = u + pv = 2864 + (8 × 10³) × 0.03432 = 3138.56 kJ/kg

• All other quantities are in kPa except pressure. So to convert pressure from MPa to kPa is first step.
• While calculating, take care of all the SI units.
• 1 bar = 10⁵ Pa
• 1 MPa = 1000 kPa = 10 bar

Concept:

Volume of vessel (V) = Volume of liquid (V_f) + Volume of vapour (V_g)

V_f = v_f × m_f

V_g = v_g × m_g

v_f,g = Specific volume of liquid and vapour

m_f,g = mass of liquid and vapour

Calculation:

Given:

V = 0.04 m³

v_f = 0.0011 m³/kg, m_f = 5 kg, v_g = 0.12 m³/kg

V_f = v_f × m_f = 0.0011 × 5 = 0.0055 m³

V = V_f + V_g = m_fv_f + m_gv_g

0.04 = 5(0.0011) + m_g(0.12)

Explanation:

Fusion is a solid to a liquid phase transition.

For water, v” < v’ (Indicating contraction) $\Rightarrow \frac{dp}{dT}\to -ve$

For most other substances v” > v’ (expansion) $\Rightarrow \frac{dp}{dT}\to +ve$

Important Points 

• The fusion curve has been drawn with a positive slope, which is typically the case.
• The fusion curve of ice/water has a negative slope due to the fact that when ice melts, the molar volume decreases.
• Ice actually melts at a lower temperature at a higher pressure

Therefore, The slope of the curve is positive as well as negative hence, it is actually variable in nature.

Explanation:

In thermodynamics, steam can exist in various states, including saturated, sub-cooled, and super-heated. These states depend on both pressure and temperature conditions.

Saturated Condition: In this state, the steam's temperature and pressure correspond to the boiling point of water. If you try to heat it more at constant pressure, it will start converting into superheated steam.

Sub-cooled Condition: Here, the steam is actually in the water state, existing at temperatures below the boiling point for a given pressure. This is more like hot water rather than steam.

Super-heated Condition: This is the state of steam where it is at a higher temperature than the saturation temperature for a given pressure. In other words, the steam has been heated above its boiling point, after all, the water content has turned to steam.

here, the steam's pressure is lower than the saturated pressure for a given temperature. This means that the steam is at a temperature higher than its boiling point for that pressure, and thus it is in a super-heated state. In superheated steam, the temperature can rise without a rise in pressure, until it reaches the new saturation pressure for that higher temperature.

Concept:

Throttling Calorimeter: A throttling calorimeter is based on the principle of throttling the wet steam so that it becomes superheated.

Separating Calorimeter: It is a vessel used initially to separate some of the moisture from the steam, to ensure superheat conditions after throttling.

Combined Separating and Throttling Calorimeter

A combined separating and throttling calorimeter is therefore found most suitable for accurate measurement of dryness of steam reducing the drawback of Separating Calorimeter (i.e. water particles from wet steam are not fully separated) and the drawback of Throttling Calorimeter (i.e. not suitable for very wet steam).

x₁ = dryness fraction of steam considering separating calorimeter

x₂ = dryness fraction of steam considering the throttling calorimeter

x = actual dryness fraction of steam in the sample

Calculation:

Actual dryness fraction of steam in the sample is:

X = x₁ × x₂

Given:             

x ₁ = 0.8, x₂ = 0.8   

Calculation:       

x = 0.8 × 0.8

Explanation:

• Adiabatic process: It is the thermodynamic process in which there is no exchange of heat and mass between the system and its surrounding during the process.
• The adiabatic process can be either reversible or irreversible.
• The essential conditions for the adiabatic process to take place:
  1. ​​The system must be perfectly insulated from the surrounding.
  2. The process must be carried out quickly so that there is no sufficient amount of time for heat and mass transfer to take place.
• ​​The adiabatic process equation:

​⇒ PVγ = constant  

The dryness fraction of wet steam undergoing adiabatic expansion can increase, decrease, or remain constant depending on several factors.

Factors affecting dryness fraction:

• Initial dryness fraction: The starting dryness fraction of the wet steam significantly impacts the outcome. If the initial dryness fraction is high (closer to 1), it's less likely to decrease significantly during expansion.
• Temperature and pressure changes: During adiabatic expansion, the temperature and pressure of the steam change. The specific direction and magnitude of these changes influence the phase change behavior of the water in the steam. Depending on the temperature and pressure drop, some of the liquid water might evaporate, increasing the dryness fraction, or some of the vapor might condense, decreasing the dryness fraction.
• Friction: While adiabatic expansion implies no heat transfer, internal friction within the steam can generate heat locally. This heat can influence the phase change behavior, potentially promoting some evaporation and increasing the dryness fraction.
• Specific heat capacities: The specific heat capacities of steam and water also play a role. If the temperature drop during expansion is more prominent for the liquid water component than the vapor component, it can result in less condensation and a potential increase in dryness fraction.