Properties of Ideal Gases and Pure Substances
Extensive and Intensive Properties
Learning Objectives:

Define Intensive and Extensive properties.

Describe the hypothetical substance ’ideal gas’ and the idealgas equation of state.

Apply the ideal gas equation of state in the solution of typical problems.
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).
Solved Example: 69101
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
Correct Answer: A
Solved Example: 69102
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: 69103
Temperature of a gas is produced due to: (SSC JE ME March 2017)
A. Its heating value
B. Kinetic energy of molecules
C. Repulsion of molecules
D. Attraction of molecules
Correct Answer: B
Solved Example: 69104
An open system is one in which: (Based on Vizag MT Mech 2015)
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
Correct Answer: C
Solved Example: 69105
A closed system is one in which: (SJVNL JE Mech 2018)
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
Correct Answer: A
Solved Example: 69106
Mixture of ice and water form a:
A. Closed system
B. Open system
C. Isolated system
D. Heterogeneous system
Correct Answer: D
Solved Example: 69107
Which of the following are the properties of a system: (ISRO VSSC Tech Asst Mech Feb 2015)
A. Pressure and temperature
B. Volume and density
C. Enthalpy and entropy
D. All of the above
Correct Answer: D
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. \[\dfrac{V_1}{T_1} = \dfrac{V_2}{T_2}\]
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:
\[PV = nRT\]
Van Der Waals Equation:
The van der Waals equation corrects for the volume of, and attractive forces between, gas molecules:
\[(P + \dfrac{a}{\bar{v}^2})(\bar{v} b) = \bar{R}T\]
There are two corrective factors in van der Waals equation. The first, \(\dfrac{a}{\bar{v}^2}\) , alters the pressure in the ideal gas equation. It accounts for the intermolecular attractive forces between gas molecules.
The factor b accounts for the volume occupied by the gas molecules. Since b corresponds to the total volume per mole occupied by gas molecules, it closely corresponds to the volume per mole of the liquid state, whose molecules are closely layered. b is generally much smaller in magnitude than a. The values of a and b generally increase with the size and complexity of the molecule.
Avogadro’s Hypothesis:
Equal volumes of gas contain equal numbers of molecules.
Solved Example: 69201
A perfect gas at 27$^\circ$C is heated at constant pressure till its volume is double. The final temperature is: (Based on CIL MT Mech 2020)
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 \[P_{1}=P_{2}\] \[\dfrac {V_{1}}{T_{1}}=\dfrac {V_{2}}{T_{2}}\] \[T_{2} =\dfrac {V_{2}T_{1}}{V_{1}} =\dfrac {2V_1 \times 300}{V_{1}} =300 \times 2=600\ K =327^{\circ}\]
Correct Answer: B
Solved Example: 69202
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
Correct Answer: A
Solved Example: 69203
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: 69204
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
Correct Answer: A
Solved Example: 69205
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
Correct Answer: A
Solved Example: 69206
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. Nonreacting 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.
Correct Answer: D
Solved Example: 69207
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. 83.14J/kmol K
C. 848J/kmol K
D. All J/kmol K
Correct Answer: B
Solved Example: 69208
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
Correct Answer: A
Specific Heats
Learning Objectives:

Establish the formula to calculate specific heat capacity.

Calculate specific heat capacity.
Heat Capacity at Constant Pressure,
\[c_p = \left(\dfrac{\partial h}{\partial T}\right)_P\]
Heat Capacity at Constant Volume,
\[c_v = \left(\dfrac{\partial u}{\partial T}\right)_v\]
where,
u = Specific Internal Energy = \(\dfrac{U}{m}\)
h = Specific Enthalpy = u + Pv = \(\dfrac{H}{m}\) For ideal gases,
\[c_p – c_v = R\]
where,
R = gas constant
Solved Example: 69301
Solids and liquids have: (UPSSSC JE Mech 2015)
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: 69302
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
Correct Answer: B
Solved Example: 69303
The ratio of two specific heats of air is equal to: (ISRO SAC Tech Asst Mech July 2018)
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: 69304
Universal gas constant is defined as equal to product of the molecular weight of the gas and: (TANGEDCO AE EC 2018)
A. Specific heat at constant pressure
B. Specific heat at constant volume
C. Ratio of two specific heats
D. Gas constant
Correct Answer: D
Solved Example: 69305
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: (ISRO RAC 2018)
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, $V = 2.5 \times 3 \times 3 = 22.5 m^3$
\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(T20)\\
T &= 470\ ^\circ C
\end{align*}
Correct Answer: D
Enthalpy
Learning Objectives:

Define enthalpy.

Properly express the enthalpy change of chemical reactions.
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: 69401
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
Entropy
Learning Objectives:

Relate the reversibility or irreversibility of a process to its change in entropy.
Entropy is a measure of molecular disorder.
When a gas expands into a vacuum, its entropy increases because the increased volume allows for greater atomic or molecular disorder. The greater the number of atoms or molecules in the gas, the greater the disorder. The magnitude of the entropy of a system depends on the number of microscopic states, or microstates, associated with it (in this case, the number of atoms or molecules); that is, the greater the number of microstates, the greater the entropy.
Chemical and physical changes in a system may be accompanied by either an increase or a decrease in the disorder of the system, corresponding to an increase in entropy (\(\Delta\)S \(>\) 0) or a decrease in entropy (\(\Delta\)S \(<\) 0), respectively. As with any other state function, the change in entropy is defined as the difference between the entropies of the final and initial states: \[\Delta S = S_f  S_i\]
Solved Example: 69501
Entropy change depends on:
A. Heat transfer
B. Mass transfer
C. Change of temperature
D. Thermodynamic state
Correct Answer: A
Solved Example: 69502
Entropy of the universe is: (DSSSB JE ME 2019 Shift I  Nov 2019)
A. Constant
B. Continuously decreasing
C. Continuously increasing
D. Considered zero if the temperature is less than 0$^\circ$C
Correct Answer: C
Solved Example: 69503
A process CANNOT be spontaneous (productfavored) 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: 69504
The occurrence of a reaction is impossible if:
A. $\Delta$H> 0; $\Delta$S >0 but $\Delta$H
B. $\Delta$H<0 ; $\Delta$S < 0 but $\Delta$H>T$\Delta$S
C. $\Delta$H <0; $\Delta$S >0
D. $\Delta$H>0 ; $\Delta$S <0
Correct Answer: D
Solved Example: 69505
In adiabatic expansion of a system in which its temperature changes from a value $T_1$ to $T_2$ the entropy will: (RSMSSB Lab Assistant 2016)
A. Increase
B. Decrease
C. Remain unchanged
D. May increase or decrease depending upon the ratio $\dfrac{T_1}{T_2}$
Correct Answer: C
Gibbs Free Energy
Learning Objectives:

To gain an understanding of Gibbs free energy.

To understand the relationship between the sign of Gibbs free energy change and the spontaneity of a process.

To be able to determine Gibbs free energy using standard free energies of formation.
Gibbs Free Energy is a thermodynamic potential that can be used to calculate the maximum of reversible work that may be performed by a thermodynamic system at a constant temperature and pressure (isothermal, isobaric). The Gibbs free energy (J in SI units) is the maximum amount of nonexpansion work that can be extracted from a thermodynamically closed system (one that can exchange heat and work with its surroundings, but not matter); this maximum can be attained only in a completely reversible process. When a system transforms reversibly from an initial state to a final state, the decrease in Gibbs free energy equals the work done by the system to its surroundings, minus the work of the pressure forces.
\[G = H  Ts\] Here
H = Enthalpy,
T= Temperature,
S= Entropy of the system
The Gibbs free energy of the system is a state function because it is defined in terms of thermodynamic properties that are state functions. The change in the Gibbs free energy of the system that occurs during a reaction is therefore equal to the change in the enthalpy of the system minus the change in the product of the temperature times the entropy of the system.
Solved Example: 69601
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: 69602
The relation between $\Delta$G and $\Delta$H is: (Based on GATE Chemistry 2020)
A. $\Delta$H = $\Delta$G  T$\Delta$S
B. $\Delta$G = $\Delta$H  T$\Delta$S
C. T$\Delta$S  $\Delta$G = $\Delta$H
D. $\Delta$H = T$\Delta$G + $\Delta$S
Gibbs Free Energy ($\Delta$G) is the energy that is available to do useful work.
It is a thermodynamic potential that can be used to calculate the maximum of reversible work that may be performed by a thermodynamic system at a constant temperature and pressure (isothermal, isobaric).
The Gibbs free energy
\[\Delta G = \Delta H  T\Delta S\]
is the maximum amount of nonexpansion work that can be extracted from a thermodynamically closed system.
Correct Answer: B
Helmholtz Free Energy
Learning Objectives:

Define Helmholtz free energy, internal energy and specific entropy.

Establish expressions for these functions given an equation of state for a perfect fluid.
Helmholtz free energy is a thermodynamic potential that measures the "useful" work obtainable from a closed thermodynamic system at a constant temperature and volume. The negative of the difference in the Helmholtz energy is equal to the maximum amount of work that the system can perform in a thermodynamic process in which volume is held constant. If the volume is not held constant, part of this work will be performed as boundary work. The Helmholtz energy is commonly used for systems held at constant volume. Since in this case no work is performed on the environment, the drop in the Helmholtz energy is equal to the maximum amount of useful work that can be extracted from the system. For a system at constant temperature and volume, the Helmholtz energy is minimized at equilibrium.
The Helmholtz free energy (F) can be defined as, \[F = U  TS\] Here
U = Internal energy,
T= Temperature,
S= Entropy of the system
So Helmholtz free energy is the amount of energy requires for creating a system by spontaneous transfer of energy to the system from the environment.
Solved Example: 69701
During a dynamite explosion, the internal energy changes from 0 J to 35000 J and entropy began at 7KJ/K and changed to 12KJ/K while temperature remained constant at 150$^\circ$C. Calculate the Helmholtz Free Energy for explosion.
A. 1557 kJ
B. 2086 kJ
C. 2168 kJ
D. 2589 kJ
T$_K$= 150 $^\circ$ C + 273.15 = 423.315 K
$\Delta$ U = 35000J  0 J= 35,000 J
$\Delta$ S = 12 KJ/K  7 KJ/K = 5 KJ / K
Since
\begin{align*}
\Delta F = \Delta U  T\Delta S &= 35000J  (423.315 K \times 5 KJ/K)\\
&= 35000J  (423.315 K \times 5000 J/K)\\
&= 2086575 J = 2086.575 kJ
\end{align*}
Correct Answer: B
Solved Example: 69702
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 kJ
\end{align*}
Correct Answer: D
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 subcooled 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: 69801
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
Correct Answer: D
Solved Example: 69802
The latent heat of vaporization at the critical point is: (SSC JE ME Paper 4 Jan 2018 Morning)
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: 69803
In a pVT surface, the zone below the triple point is known as:
A. Liquid zone
B. Vapor zone
C. Sublimation zone
D. None of these
Correct Answer: A
Solved Example: 69804
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: (GATE ME 2015)
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: 69805
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: 69806
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$^2$
Correct Answer: D
Properties of Steam
Learning Objectives:

Use the Mollier Diagram to estimate the properties of steam.

Use steam tables.
The ratio of the mass of the mixture which is vapor (vap) to the total mixture mass is called quality of steam.
\[x = \dfrac{m_{\mathrm{vap}}}{m_{\mathrm{total}}}\]

x = 0, corresponds to \(m_{vap}\) = 0. This is the all liquid limit.

x = 1, corresponds to \(m_{vap}\) = \(m_{total}\) . 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, \[u_g = u_f + u_{fg}\] Enthalpy, \[h_g = h_f + h_{fg}\] Entropy, \[s_g = s_f + s_{fg}\]
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.
\[h = h_f + v_f(P  P_{sat})\]
Solved Example: 69901
Superheated vapour behaves:
A. Exactly as gas
B. As steam
C. As ordinary vapour
D. Approximately as a gas
Correct Answer: D
Solved Example: 69902
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. (SSC JE ME Sep 2019 Evening)
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\\ s_f &= 2.794\ KJ/kg K\\ s_{fg} &= 3.277\ KJ/kg K. \end{align*} Enthalpy of mixture, \[h = h_f + x h_{fg} = 1085.8 + 0.06 \times 1714.6 = 1188.67\ KJ / kg\] Entropy of mixture, \[s = s_f + x s_{fg} = 2.794 + 0.06 \times 3.277 = 2.99\ kJ/kg K\] Internal energy, \[u = h  p v = 1188.67  (39.776 \times 10^2 \times 0.00418) = 1172\ KJ/kg\]
Correct Answer: D
Solved Example: 69903
Gaseous mixtures _________.
A. Can only contain isolated atoms
B. Can only contain molecules
C. Are all heterogeneous
D. Are all homogeneous
Correct Answer: D
Solved Example: 69904
An ideal gas as compared to a real gas at very high pressure occupies: (SJVNL JE Mech 2018)
A. More volume
B. Less volume
C. Same volume
D. Unpredictable behaviour
Correct Answer: A
Solved Example: 69905
Which of the following can be regarded as gas so that gas laws could be applicable, within the commonly encountered temperature limits? (SSC JE ME March 2017 Morning)
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
Correct Answer: D
Solved Example: 69906
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
Correct Answer: C
Solved Example: 69907
No liquid can exist as liquid at:
A. 273 K
B. Vacuum
C. Zero pressure
D. Centre of earth
Correct Answer: C
Solved Example: 69908
The condition of perfect vacuum, i.e., absolute zero pressure can be attained at: (ISRO RAC 2014)
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
Correct Answer: A
Solved Example: 69909
The same volume of all gases would represent their:
A. Densities
B. Specific weights
C. Molecular weights
D. Gas characteristic constants
Correct Answer: C
Solved Example: 69910
What is standard temperature and pressure (STP)? (SSC Scientific Asst Nov 2017Shift II)
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
Correct Answer: D