Properties of Ideal Gases and Pure Substances
Thermodynamic Properties
Learning Objectives:
- Define intensive and extensive properties.
- Establish the formula to calculate specific heat capacity.
- Define enthalpy.
Extensive and Intensive Properties:
Extensive properties are proportional to mass. Extensive properties depend on the amount of material. These are a material’s properties such as mass and volume.
Intensive properties are independent of mass. These are either properties like temperature or others that combine, or are the ratio of, two extensive variables like density (mass/volume) or molar volume (volume/mole).
State Functions (Properties):
- Specific Volume = v = $\dfrac{V}{m}$
- Specifc Internal Energy = u = $\dfrac{U}{m}$
- Specific Enthalpy = h = $\dfrac{H}{m}$ = $\dfrac{U + PV}{m}$ = h + Pv
- Specific Entropy = s = $\dfrac{S}{m}$
- Gibbs Free Energy G = $h - Ts$
- Helmholtz Free Energy A = $u - Ts$
Specific Heats:
Heat Capacity at Constant Pressure,
Heat Capacity at Constant Volume,
where,
u = Specific Internal Energy = \(\dfrac{U}{m}\)
h = Specific Enthalpy = u + Pv = \(\dfrac{H}{m}\) For ideal gases,
where,
R = gas constant
Olivier Cleynen, CC0, via Wikimedia Commons
Enthalpy:
The sum of the internal energy of the system plus the product of the pressure of the gas in the system and its volume.
Thermodynamic Definition of Enthalpy (H): \[H = U + PV\] U = energy of the system
P = pressure of the system
V = volume of the system
Properties of Enthalpy:
When calculating enthalpy changes associated with chemical reactions we will make use of several properties of enthalpy.
- Enthalpy is a state function. Therefore, we can determine the enthalpy change, \(\Delta\)H, of a chemical reaction by subtracting the enthalpy of the reactants from the enthalpy of the products:
\(\Delta H_{reaction}\) = \(\Sigma H_{products}\) - \(\Sigma H_{reactants}\) - Enthalpy is an extensive property.
Solved Example: 69-1-01
Extensive property of a system is one whose value:
A. Depends on the mass of the system like volume
B. Does not depend on the mass of the system, like temperature, pressure, etc.
C. Is not dependent on the path followed but on the state
D. Is dependent on the path followed and not on the state
Extensive properties depend upon the substance present in the system. More substance will give more property. An exampe will be moass and volume.
Intensive properties do not depend upon how much substance present in the system. An example will be density. Just because you take more substance in the container does not increase the density of the material.
Correct Answer: A
Solved Example: 69-1-02
Which of the following variables controls the physical properties of a perfect gas?
A. Pressure
B. Temperature
C. Volume
D. All of the above
Correct Answer: D
Solved Example: 69-1-03
Temperature of a gas is produced due to:
A. Its heating value
B. Kinetic energy of molecules
C. Repulsion of molecules
D. Attraction of molecules
Temperature is caused by the motion of molecules which gives them Kinetic energy. At 0K, there is no KE and there is no temperature.
Correct Answer: B
Solved Example: 69-1-04
An open system is one in which:
A. Mass does not cross boundaries of the system, though energy may do so
B. Neither mass nor energy crosses the boundaries of the system
C. Both energy and mass cross the boundaries of the system
D. Mass crosses the boundary but not the energy
In closed system, the matter cannot transfer outside the control volume, whereas in open system, the mass transfer in and out of the control volume. Energy transfer in both cases is possible.
In isolated system, the system does not transfer mass as well as energy with surroundings.
Correct Answer: C
Solved Example: 69-1-05
A closed system is one in which:
A. Mass does not cross boundaries of the system, though energy may do so
B. Mass crosses the boundary but not the energy
C. Neither mass nor energy crosses the boundaries of the system
D. Both energy and mass cross the boundaries of the system
In closed systems, mass does not cross the system boundaries, but heat and work can cross the system boundaries.
Correct Answer: A
Solved Example: 69-1-06
Mixture of ice and water form a:
A. Closed system
B. Open system
C. Isolated system
D. Heterogeneous system
While the chemical formula for ice and water is same, the phases (physical composition) of ice and water are different. Ice is in solid phase as compared to water is in liquid phase. Hence, they form a heterogeneous system.
Correct Answer: D
Solved Example: 69-1-07
Which of the following are the properties of a system:
A. Pressure and temperature
B. Volume and density
C. Enthalpy and entropy
D. All of the above
All measurable characteristics are properties. Here, pressure, temperature, volume, density, enthalpy and entropy all are measurable and hence are properties. Properties are further divided into two categories such as extensive and intensive properties.
Correct Answer: D
Solved Example: 69-3-01
Solids and liquids have:
A. One value of specific heat
B. Two values of specific heat
C. Three values of specific heat
D. No value of specific heat
Since solids and liquids do not expand significantly when heated, they have only one specific heat. However, gases have two specific heats depending upon whether the expansion is allowed (constant pressure process) or prevented (constant volume process).
Correct Answer: A
Solved Example: 69-3-02
Gases have:
A. Only one value of specific heat
B. Two values of specific heat
C. Three values of specific heat
D. No value of specific heat
Gases have two values of specific heat.
- Specific Heat under constant volume (C$_v$)
- Specific Heat under constant pressure (C$_p$)
Correct Answer: B
Solved Example: 69-3-03
The ratio of two specific heats of air is equal to:
A. 0.17
B. 0.24
C. 0.1
D. 1.41
The ratio $\dfrac{C_p}{C_v} = \gamma$ depends upon whether how many atoms are there in molecules. For monoatomic gas $\gamma$ = 1.67, for diatomic gas, $\gamma$ = 1.4 and for triatomic gas $\gamma$ = 1.33. Since major composition of air is made by diatomic gases ($N_2$ and $O_2$) air is considered diatomic and its $\gamma$ value is 1.4
Correct Answer: D
Solved Example: 69-3-04
Universal gas constant is defined as equal to product of the molecular weight of the gas and:
A. Specific heat at constant pressure
B. Specific heat at constant volume
C. Ratio of two specific heats
D. Gas constant
\begin{align*} R &= \dfrac{\bar{R}}{\mathrm{mol.\ wt.}}\\ \bar{R} &= {\mathrm{mol.\ wt.}} \times R \end{align*}
Correct Answer: D
Solved Example: 69-3-05
A 100 W electric bulb was switched on in a 2.5 m $\times$ 3m $\times$ 3m size thermally insulated room having a temperature of 20$^\circ$C. The room temperature at the end of 24 hours will be:
A. 321
B. 341
C. 450
D. 470
\[\mathrm{Heat\ generated\ by\ bulb} = \mathrm{Power} \times \mathrm{time}\]
\begin{align*}
&= 100 \times 24 \times 60\times 60\\
&= 8.64 \times 10^6 J
\end{align*}
Density of the air, $\rho$ = 1.2 kg/m$^3$
C$_v$ = 0.718 KJ/Kg K
Volume of the room, $V = 2.5 \times 3 \times 3 = 22.5\ \mathrm{m}^3$
\begin{align*}
\mathrm{Energy} &= m C_v \Delta T\\
&= \rho V C_v (T -20)\\
8.64 \times 10^6 &= 1.2 \times 22.5 \times 0.718 \times 10^3 \times(T-20)\\
T &= 470\ ^\circ C
\end{align*}
Correct Answer: D
Solved Example: 69-4-01
Total heat of a substance is also known as:
A. Internal energy
B. Entropy
C. Thermal capacity
D. Enthalpy
Enthalpy is a measurement of energy in a thermodynamic system. It is the thermodynamic quantity equivalent of the total heat content of a system. Enthalpy can be found out by adding internal energy with product of pressure and volume.
Correct Answer: D
Solved Example: 69-4-02
Enthalpy is calculated as the:
A. Sum of internal energy and the product of pressure and volume of the system
B. Sum of internal energy and the product of pressure and density of the system
C. Difference between the internal energy and the product of pressure and density of the system
D. Difference between the internal energy and the product of pressure and volume of the system
Enthalpy is calculated as:
\[H = U + pv\]
Where H =Enthalpy
U = Internal Energy
p= pressure
v = volume
Correct Answer: A
Solved Example: 69-4-03
Consider the following statement:
- In an ideal gas, there are no intermolecular forces of attraction and repulsion.
- At very low pressure, all gases and vapours approach ideal gas behavior.
- Enthalpy of an ideal gas depends only on temperature.
A. 1, 2 and 3
B. 1 and 2 only
C. 1 and 3 only
D. 2 and 3 only
Let us see each option individually.
- TRUE. There are two main postulates of the kinetic gas theory.
- There are no intermolecular forces of attractions.
- The molecules occupy no volume.
- TRUE. At very low pressure and high temperature, all gases and vapour approach to ideal gas behaviour.
- TRUE. The internal energy and enthalpy of ideal gases depends only on temperature, and hence they are STATE functions. As opposed to that work is dependent on the path taken, and hence it is a PATH function.
Correct Answer: A
Solved Example: 69-4-04
In an isothermal process, the internal energy:
A. Increases
B. Decreases
C. Remains constant
D. First increases then decreases
Correct Answer: C
Solved Example: 69-4-05
What parameter will remain constant in a throttling process?
A. Entropy
B. Temperature
C. Pressure
D. Enthalpy
Let's see each option individually.
- In isentropic process, entropy remains constant.
- In isothermal process, temperature remains constant.
- In isobaric process, pressure remains constant.
- In throttling process, enthalpy remains constant.
Correct Answer: D
Solved Example: 69-4-06
For an exothermic reaction the change in enthalpy is:
A. Positive
B. Neutral
C. Negative
D. Constant
- For exothermic reactions, the change in enthalpy is negative $\Delta$H < 0 and heat is released during reaction.
- For endothermic reactions, the change in enthalpy is positive $\Delta$H > 0 and heat is absorbed during reaction.
Correct Answer: C
Solved Example: 69-4-07
S.I. unit of enthalpy (h) is:
A. J/m
B. J/K
C. J/kg
D. K/J
Enthalpy is the total heat content of the system. It is equal to the internal energy of the system plus the product of pressure and volume. While unit of enthalpy is Joule (J), unit for specific enthalpy (h) is J/kg.
Correct Answer: C
Solved Example: 69-4-08
The mean specific heat (Cp$_{\mathrm{mean}}$) for superheated steam at 0.5 bar, between 300$^\circ$C and 400$^\circ$C, is: (Consider h$_{\mathrm{sup}}$ = 3075.5 kJ/kg at 300$^\circ$C and h$_{\mathrm{sup}}$ = 3278.9 kJ/kg at 400$^\circ$C)
A. 1.49 kJ/kg$^\circ$C
B. 0.03 kJ/kg$^\circ$C
C. 2.03 kJ/kg$^\circ$C
D. 0.49 kJ/kg$^\circ$C
\begin{align*} \Delta h &= (c_p)_{\mathrm{mean}} \times (\Delta T)\\ (3278.9 - 3075.5) &= (c_p)_{\mathrm{mean}} \times (400- 300)\\ 203.4 &= (c_p)_{\mathrm{mean}} \times (100)\\ (c_p)_{\mathrm{mean}} &= 2.034 kJ/kg^\circ C \end{align*}
Correct Answer: C
Solved Example: 69-4-09
For an exothermic reaction the change in enthalpy is:
A. Positive
B. Neutral
C. Negative
D. Constant
A chemical reaction or physical change is exothermic if heat is released by the system into the surroundings. Because the surroundings are gaining heat from the system, the temperature of the surroundings increases. The sign of $\Delta$ H for an exothermic process is negative because the system is losing heat.
Correct Answer: C
Solved Example: 69-5-03
A process CANNOT be spontaneous (product-favored) if:
A. It is endothermic, and there is a decrease in disorder.
B. It is endothermic, and there is an increase in disorder.
C. It is exothermic, and there is an increase in disorder.
D. It is exothermic, and there is a decrease in disorder.
Correct Answer: A
Solved Example: 69-6-01
What is the standard free energy change $\Delta$G for the following reaction at 25$^\circ$C? \[N_2 + 3H_2 \rightarrow 2NH_3\] The values for $\Delta$H and S is -91.8KJ and -198.0J/K
A. -32.8 KJ
B. -43.9 KJ
C. -67.0 KJ
D. -112.3 KJ
\[\Delta G = -91.8 - (298 \times -0.1980) = -32.8\ KJ\]
Correct Answer: A
Solved Example: 69-7-01
During a dynamite explosion, the internal energy changes from 0 J to 35000 J and entropy began at 7 KJ/K and changed to 12 KJ/K while temperature remained constant at 150$^\circ$C. Calculate the Helmholtz Free Energy for explosion.
A. 1557 kJ
B. 2080 kJ
C. 2168 kJ
D. 2589 kJ
T$_K$= 150 $^\circ$ C + 273 = 423 K
$\Delta$ U = 35000J - 0 J= 35,000 J
$\Delta$ S = 12 KJ/K - 7 KJ/K = 5 KJ/K
Since
\begin{align*}
A = \Delta U - T\Delta S &= 35000J - (423 K \times 5 KJ/K)\\
&= 35000J - (423 K \times 5000 J/K)\\
&= 2080\ \mathrm{kJ}
\end{align*}
Correct Answer: B
Solved Example: 69-7-02
For a thermodynamic system, change in entropy is 18KJ/K and change in internal energy is 23000 J at 250 K temperature. Calculate the Helmholtz Free Energy for the given system in KJ.
A. -2555
B. -3466
C. -3900
D. -4477
T$_K$= 250 K
$\Delta$ U = 23000 J
$\Delta$ S = 18 KJ/K
Since
\begin{align*}
\Delta F &= \Delta U - T \Delta S\\ &= 23000J - (250 K \times 18KJ/K)\\
&= 23000J - (250 K \times 18000 J/K)\\
&= - 4477\ \mathrm{kJ}
\end{align*}
Correct Answer: D
Solved Example: 69-7-05
The term (u - T.s) in deriving the maximum work done during the nonflow reversible process is known as:
A. Van-der function
B. Gibbs function
C. Helmholtz function
D. Rankine function
- Gibbs free energy $G = h - Ts$
- Helmholtz free energy $A = u - Ts$
Correct Answer: C
Ideal Gas Laws
Learning Objectives:
- Explain and apply Boyle’s, Charles’, and Avogadro’s gas Laws to observations of gas behavior.
- Apply gas laws to stoichiometric problems.
- Define the conditions of STP.
Charles’ Law:
For a fixed mass of gas at constant pressure, the volume is directly proportional to the kelvin temperature. \[\dfrac{V_1}{T_1} = \dfrac{V_2}{T_2}\]
NASA's Glenn Research Center, Public domain, via Wikimedia Commons
Boyle’s Law:
For a fixed mass of gas at constant temperature, the volume is inversely proportional to the pressure. \[p_1 V_1 = p_2 V_2\]
General Gas Equation:
Solved Example: 69-2-01
A perfect gas at 27$^\circ$C is heated at constant pressure till its volume is doubled. The final temperature is:
A. 54$^\circ$C
B. 327$^\circ$C
C. 108$^\circ$C
D. 654$^\circ$C
\[T_{1}=27^{\circ}C=27+273=300\ K\] \[V_{2}=2V_{1}\] \[\dfrac {P_{1}V_{1}}{T_{1}}=\dfrac {P_{2}V_{2}}{T_{2}}\] Given \[P_{1}=P_{2}\] \[\dfrac {V_{1}}{T_{1}}=\dfrac {V_{2}}{T_{2}}\] \begin{align*} T_{2} &=\dfrac {V_{2}T_{1}}{V_{1}}\\ &=\dfrac {2V_1 \times 300}{V_{1}}\\ &=300 \times 2\\ &=600\ K =327^{\circ}C \end{align*}
Correct Answer: B
Solved Example: 69-2-02
The statement that molecular weights of all gases occupy the same volume is known as:
A. Avogadro's hypothesis
B. Dalton's law
C. Gas law
D. Law of thermodynamics
Let's see each option individually.
- Avogadro's hypothesis states that equal volumes of all gases at the same temperature and pressure contain equal numbers of particles.
- Dalton's law of partial pressure states that the total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases.
- Gas law: There is Charle's law and Boyle's law, both of which can be combined into Ideal gas law. \[\dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2}\]
- Laws of Thermodynamics There are three laws of thermodynamics (in addition to the zeroth law).
Correct Answer: A
Solved Example: 69-2-03
Of the following, _________ is a valid statement of Charles' law.
A. VT = constant
B. V = constant
C. PT = constant
D. PV = constant
Correct Answer: A
Solved Example: 69-2-04
According to Avogadro's Hypothesis:
A. The molecular weights of all the perfect gases occupy the same volume under same conditions of pressure and temperature
B. The sum of partial pressure of mixture of two gases is sum of the two
C. Product of the gas constant and the molecular weight of an ideal gas is constant
D. Gases have two values of specific heat
Option (A) is Avogadro's hypothesis. Option (B) is Dalton's law of partial pressure.
Correct Answer: A
Solved Example: 69-2-05
Of the following, which is a correct statement of Boyle's law?
A. PV = constant
B. VT = constant
C. V/P = constant
D. PV$^n$ = constant
In Charle's law, the pressure remains constant and hence does NOT apprear in the equation.
In Boyle's law, the temperature remains constant and hence does NOT appear in the equation.
Correct Answer: A
Solved Example: 69-2-06
Which of the following statements about gases is false?
A. Distances between molecules of gas are very large compared to bond distances within molecules.
B. Non-reacting gas mixtures are homogeneous.
C. Gases expand spontaneously to fill the container they are placed in.
D. All gases are colorless and odorless at room temperature.
While most gases are colorless, there are a few exceptions:
- Fluorine, F$_2$, and Chlorine, Cl$_2$, which are pale yellow-green.
- Bromine, Br$_2$, and Nitrogen Dioxide, NO$_2$, which are reddish brown.
- Iodine, I$_2$, which is violet.
Correct Answer: D
Solved Example: 69-2-07
The value of the product of molecular weight and the gas characteristic constant for all the gases in S.I. units is:
A. 29.27 J/kmol K
B. 8314 J/kmol K
C. 848 J/kmol K
D. All of the above
The universal gas constant or molar constant (R$_u$) of a gas is the product of the specific gas constant (R) and the molecular mass of the gas (M).
R$_u$ = 8.314 kJ/kg mol K = 8314 J/kmol K
Correct Answer: B
Solved Example: 69-2-08
Molecular volume of any perfect gas at 600 $\times$ $10^3$ N/m$^2$ and 27$^\circ$C will be:
A. 4.17m$^3$/kgmol
B. 400 m$^3$/kg mol
C. 0.15 m$^3$/kg mol
D. 41.7 m$^3$/kg mol
\begin{align*} PV &= nRT\\ 600 \times 10^3 \times V &= 1 \times 8314 \times 300\\ V &= \dfrac{8314 \times 300}{600 \times 10^3}\\ &= 4.157 m^3/\mathrm{kgmol} \end{align*}
Correct Answer: A
Properties of Pure Substances
Learning Objectives:
- Define a pure substance and a phase.
-
Melting: Solid to liquid phase change
-
Sublimation: Solid to vapor phase change
-
Evaporation: Liquid to vapor phase change
-
Condensation: Vapor to liquid phase change
-
Triple point: The only state at which the solid, liquid and vapour phases coexist in equilibrium.
-
Critical point: The limit of distinction between a liquid and vapour.
-
Critical pressure: The pressure at the critical point.
-
Critical temperature: The temperature at the critical point.
-
Gas: A vapor whose temperature is greater than the critical temperature.
-
Saturation temperature: The phase change temperature corresponding to the saturation pressure. Sometimes called the boiling temperature.
-
Saturation pressure: The phase change pressure.
-
Compressed liquid. Liquid whose temperature is lower than the saturation temperature. Also called a sub-cooled liquid.
-
Sensible heat of water (h\(_f\)): It is defined as the quantity of heat absorbed by 1 kg of water when it is heated from 0\(^\circ\)C (freezing point) to boiling point.
-
Latent heat or hidden heat (h\(_{fg}\)). It is the amount of heat required to convert water at a given temperature and pressure into steam at the same temperature and pressure.
-
Dryness fraction (x): The term dryness fraction is related with wet steam. It is defined as the ratio of the mass of actual dry steam to the mass of steam containing it.
-
Saturated liquid: Liquid at the saturation temperature corresponding to the saturation pressure. That is liquid about to commence evaporating, represented by the point f on a diagram.
-
Wet vapor: The mixture of saturated liquid and dry vapor during the phase change.
-
Total heat or enthalpy of wet steam (h): It is defined as the quantity of heat required to convert 1 kg of water at 0\(^\circ\)C into wet steam at constant pressure.
Solved Example: 69-8-01
Triple point of a pure substance is a point at which:
A. Liquid and vapor exist together
B. Solid and vapor exist together
C. Solid and liquid exist together
D. Solid, liquid and vapor exist together
Triple point is that point where solid, liquid, and vapor phases coexist.
Correct Answer: D
Solved Example: 69-8-02
The latent heat of vaporization at the critical point is:
A. Equal to zero
B. Less than zero
C. Greater than zero
D. None of these
At the critical point, only one phase exists. Therefore, the heat of vaporization is zero.
Correct Answer: A
Solved Example: 69-8-03
In a p-V-T surface, the zone below the triple point is known as:
A. Liquid zone
B. Vapor zone
C. Sublimation zone
D. None of these
In a p-V-T surface, the zone immediately below the triple point is liquid zone.
Correct Answer: A
Solved Example: 69-8-04
A rigid container of volume 0.5 $m^3$ contains 1.0 kg of water at 120 $^\circ$C ($v_f$ = 0.00106 $m^3/kg$, $v_g$ = 0.8908 $m^3/kg$). The state of water is:
A. Compressed liquid
B. Saturated liquid
C. A mixture of saturated liquid and saturated vapor
D. Superheated vapor
\[v = \dfrac{0.5}{1} = 0.5 m^3/kg\] Since $v_f$ < v < $v_g$ the state of water is mixture of saturated water and saturated vapour.
Correct Answer: C
Solved Example: 69-8-05
At critical point, i.e. p=225.65 kg/cm$^2$, the latent enthalpy of vaporization is:
A. Maximum
B. Minimum
C. Zero
D. Depends on temperature also
Correct Answer: C
Solved Example: 69-8-06
At which pressure the properties of water and steam become identical?
A. 0.1 kg/cm$^2$
B. 1 kg/cm$^2$
C. 100 kg/cm$^2$
D. 225.6 kg/cm$^2$
At the critical point, the saturated liquid and saturated vapour state are identical.
For water, pressure at the critical point = 221.2 bar (225.5 kg/cm$^2$)
Correct Answer: D
Properties of Steam
Learning Objectives:
- Use the Mollier Diagram to estimate the properties of steam.
- Use steam tables.
Quality of Steam:
The ratio of the mass of the mixture which is vapor (vap) to the total mixture mass is called quality of steam.
m$_g$ = mass of vapor
m$_f$ = mass of liquid
- x = 0, corresponds to $m_{g}$ = 0. This is the all liquid limit.
- x = 1, corresponds to $m_{f}$ = 0. This is the all gas limit. \[0\leq x\leq 1\]
Saturation pressure is the pressure at which the liquid and vapor phases are in equilibrium at a given temperature.
Saturation Temperature is the temperature at which the liquid and vapor phases are in equilibrium at a given pressure.
Internal Energy,
Enthalpy,
Entropy,
Superheated Steam Properties:
- Saturated vapor: A term including wet and dry vapor.
- Dry (saturated) vapor. Vapor which has just completed evaporation. The pressure and temperature of the vapor are the saturation values.
- Superheated steam: When steam is heated after it has become dry and saturated, it is called superheated steam and the process of heating is called superheating. Superheating is always carried out at constant pressure.
- Degree of superheat: The term used for the numerical amount by which the temperature of a superheated vapor exceeds the saturation temperature.
Solved Example: 69-9-01
Superheated vapour behaves:
A. Exactly as gas
B. As steam
C. As ordinary vapour
D. Approximately as a gas
At higher temperature i.e. at a superheated region, a vapour approximately follows the perfect gas law.
Correct Answer: D
Solved Example: 69-9-02
A vessel of volume 0.04 m$^3$ contains a mixture of saturated water and steam at a temperature of 250$^\circ$C. The mass of the liquid present is 9 kg. Find the internal energy.
A. 563 KJ/kg
B. 777 KJ/kg
C. 972 KJ/kg
D. 1172 KJ/kg
Volume, V = 0.04 m$^3$
Temperature, T = 250$^\circ$C
Mass, m = 9 kg
From the Steam tables corresponding to 250$^\circ$C,
\begin{align*}
v_f &= v_1 = 0.001251 m^3/kg \\
v_g &= v_s = 0.050037 m^3/kg \\
p &= 39.776\ bar
\end{align*}
Total volume occupied by the liquid,
\[V_1 = m_1 \times v_1
= 9 \times 0.001251
= 0.0113\ m^3\]
Total volume of the vessel,
V = Volume of liquid + Volume of steam
\begin{align*}
V &= V_1 + V_S\\
0.4 &= 0.0113 + V_S \\
V_S &= 0.0287\ m^3.
\end{align*}
Mass of steam,
\[m_s = V_S / v_s = 0.0287 / 0.050037 = 0.574\ kg\]
Mass of mixture of liquid and steam,
\[m = m_1 + m_s = 9 + 0.574 = 9.574\ kg\]
Total specific volume of the mixture,
\[v = \dfrac{0.04}{9.574} =0.00418\ m^3 / kg\]
We know that,
\begin{align*}
v &= v_f + x vf_g\\
0.00418 &= 0.001251 + x (0.050037 - 0.001251) \\
x &= 0.06
\end{align*}
From Steam table corresponding to 250$^\circ$C,
\begin{align*}
h_f &= 1085.8\ KJ/kg \\
h_{fg} &= 1714.6\ KJ/kg\\
\end{align*}
Enthalpy of mixture,
\begin{align*}
h &= h_f + x h_{fg}\\
&= 1085.8 + 0.06 \times 1714.6\\
&= 1188.67\ KJ / kg
\end{align*}
Internal energy,
\begin{align*}
u &= h - p v \\
&= 1188.67 - (39.776 \times 10^2 \times 0.00418) \\
&= 1172\ KJ/kg
\end{align*}
Correct Answer: D
Solved Example: 69-9-03
Gaseous mixtures _________.
A. Can only contain isolated atoms
B. Can only contain molecules
C. Are all heterogeneous
D. Are all homogeneous
Gaseous mixtures are homogeneous because unlike solids and liquids, gases mix with each other, even if they do not react chemically. An example would be air, which is a homogeneous mixture of N$_2$, O$_2$, CO$_2$ and other gases in small quantity.
Correct Answer: D
Solved Example: 69-9-04
An ideal gas as compared to a real gas at very high pressure occupies:
A. More volume
B. Less volume
C. Same volume
D. Unpredictable behaviour
Real gases occupy more volume than ideal gases at high pressures. This is because the ideal gas law assumes that mass is a point that occupies no volume. The actual molecule occupies a fraction of the volume, and at high pressure this ratio becomes important. This means that molecules travel a shorter distance between collisions than they would if they had zero size. The higher the number of collisions per unit time, the higher the pressure for a given density and temperature.
Correct Answer: A
Solved Example: 69-9-05
Which of the following can be regarded as gas so that gas laws could be applicable, within the commonly encountered temperature limits?
A. O$_2$, N$_2$, steam, CO$_2$
B. O$_2$, N$_2$, water vapor
C. SO$_2$, NH$_3$, CO$_2$, moisture
D. O$_2$, N$_2$, H$_2$, air
Gases like O$_2$, N$_2$, H$_2$, and air are subject to the gas law. These gases have little to no interatomic force of attraction between the molecules, non-polarization, and a minimal molecular volume in comparison to the volume of the container.
Correct Answer: D
Solved Example: 69-9-06
Absolute zero pressure will occur:
A. At sea level
B. At the center of the earth
C. When molecular momentum of the system becomes zero
D. Under vacuum conditions
At sea level, we still have atmospheric pressure, andd the pressure increases with an increase in depth by the formula $P = h \rho g$, so options (A) and (B) are incorrect.
Pressure is exerted due to molecular motion. When this molecule motion is zero, there is no thrust exerted on the walls of the container, hence the pressure becomes zero.
Correct Answer: C
Solved Example: 69-9-07
No liquid can exist as liquid at:
A. -273 K
B. Vacuum
C. Zero pressure
D. Centre of earth
A liquid will boil and vapourize if brought to zero pressure (vacuum). Despite the fact that different liquids have varying viscosities, they eventually vapourize if there is no pressure to keep them from losing their viscosity and surface tension.
And since space is essentially a vacuum with no pressure, water will begin to boil there.
We now have some isolated water molecules in a gaseous form after the water has boiled or vapourized, but in a very cold atmosphere. These small droplets of water vapor now instantly freeze (desublimate), forming ice crystals.
Correct Answer: C
Solved Example: 69-9-08
The condition of perfect vacuum, i.e., absolute zero pressure can be attained at:
A. A temperature of -273.16$^\circ$C
B. A temperature of 0$^\circ$C
C. A temperature of 273 K
D. A negative pressure and 0$^\circ$C temperature
At absolute zero temperature, which is 0K, or -273.15$^\circ$C, according to kinetic gases theory, the intermolecular momentum is zero and we can have absolute zero pressure.
Correct Answer: A
Solved Example: 69-9-09
The same volume of all gases would represent their:
A. Densities
B. Specific weights
C. Molecular weights
D. Gas characteristic constants
One mole of any gas occupies exactly same volume (22.4 liter) at standard temperature and pressure (STP).
So same quantities of gas will be represented by same number of moles, and hence same multiples of their molecular weights.
Correct Answer: C
Solved Example: 69-9-10
What is standard temperature and pressure (STP)?
A. 0 $^\circ$F and 14.7 psia
B. 0 $^\circ$C and 1 liter volume
C. 32 $^\circ$F and 0 psia
D. 32 $^\circ$F and 1 atm
Standard Temperature and Pressure (STP) is defined as 0 $^\circ$C (which is same as 32$^\circ$F) and 1 atmosphere of pressure.
Correct Answer: D