271. The equation which perfectly represents Gibbs Phase Rule for a process where the pressure is also a variable is expressed as :
(a) F = P – C + 2
(b) F = C + P – 1
(c) F = P – C – 1
(d) F = C – P + 2
BHEL ET 2019
. (d) : Gibbs phase rule- P + F = C + 2 or F C P 2 = − + P = Number of phase C = Number of chemical components n = number of non-compositional variable (i.e. n=2) h = 2 i.e., pressure and temperature F = Degree of freedom
Answer
272. Which relationship defines Gibbs free energy G:
(a) G = H + TS
(b) G = H - TS
(c) G = U + TS
(d) F = U + TS
SJVN ET 2013
APPSC AEE 2016
. (b) : Gibbs free energy 'G' G H TS = −
Answer
273. Joule-Thomson coefficient is defined as :
(a) h T P ∂ ∂
(b) T H P ∂ ∂
(c) P H T ∂ ∂
(d) h P T ∂ ∂
MPPSC AE 2016
BPSC AE Mains 2017 Paper - V
: (a) Joule thomson co-efficient:- The change in temperature with drop in pressure at constant enthalpy is termed as Joule-Thomson coefficient (µ). h T p ∂ µ = ∂ It varies with both the temperature and pressure of the gas. ∗ Joule thomson coefficient for an ideal gas is zero.
Answer
274. The statement that molecular weight of all gases occupy the same volume is known as-
(a) Avogardo's hypothesis
(b) Gas law
(c) Dalton's law
(d) Thermodynamics law
RPSC AE 2018
. (a) : The statement that molecular weight of all gases occupy the same volume is known as Avogardo's hypothesis.
Answer
275. Which is a single phase system ?
(a) Mixture of water and alcohol
(b) Mixture of oil and water
(c) Liquid water, ice and water vapour
(d) Water and ice
OPSC Civil Services Pre. 2011
. (a) : Single phase system is mixture of water and alcohol.
Answer
276. For an ideal gas the compressibility factor is:
(a) Zero
(b) Unity
(c) Infinity
(d) None of these
OPSC AEE 2019 PAPER - II
: (b) : For an ideal gas, PV = RT i.e. the compressibility factor, z =1 PV As, Z RT =
Answer
277. A sample of ideal gas has an internal energy U and is then compressed to one-half of its original volume while the temperature stays the same. What is the new internal energy of the ideal gas in terms of U ?
(a) U
(b) 1/2U
(c) 1/4U
(d) 2U
RPSC Vice Principal ITI 2018
. (a) : Internal energy of a perfect gas is only function of temperature only. For real gas internal energy is function of temperature and specific volume both U = f (T, v).
Answer
278. Clausius-Clapeyron equation gives the slope of the curve in
(a) p-v diagram
(b) p-h diagram (c) p-T diagram
(d) T-S diagram
BPSC AE Mains 2017 Paper - V
: (c) : Clausius-Clapeyron equation, g f 2 dP (h – h ) P dT RT × = It gives slope of p – T curve.
Answer
279. Select the correct statement as per Charles's law.
(a) PV = Constant, if T is kept constant.
(b) V T = constant, if P is kept constant.
(c) P V = constant, if T is kept constant.
(d) T P = constant, if T is kept constant.
TSPSC AEE 2015
: (b) Charles's law:- The volume of a given mass of a perfect Gas varies directly as its absolute temperature, when the absolute pressure remains constant V ∝ T V costant T = 1 2 3 1 2 3 V V V T T T = = = constant
Answer
280. Which of the following is a general gas equation?
(a) PVn = C
(b) PV = C
(c) PV = RT (d) PV = mRT
UJVNL AE 2016
: (d) For any mass m kg of gas, the characteristic gas equation PV = mRT In S.I unit, the pressure is expressed in bar The unit of gas constant (R) in SI unit is N-m/kg.K R = 287 J/kg K Or 0.287 kJ/kg K The equation PV = mRT may also be expressed in another form m P RT RT V = = ρ
Answer
281. Which one of the following relations defines Helmholtz function?
(a) H + TS
(b) H – TS
(c) U + TS
(d) U – TS
OPSC AEE 2019 PAPER - II
: (d) : Helmholtz function (F), F = U – TS → It is applicable for open system. Gibbs function (G), G = H – TS → It is applicable for open system.
Answer
282. ITPS defines which of the following as a reference for calibration of temperature scale ?
(a) Ice Point
(b) Lamda Point
(c) Normal Boiling Point of Water
(d) Triple Point of Water
OPSC Civil Services Pre. 2011
. (d) : Triple point of water is a reference for calibration of temperature scale.
Answer
283. In a single-component condensed system, if degree of freedom is zero, maximum number of phases that can co-exist_____
(a) 0
(b) 1
(c) 2
(d) 3
OPSC AEE 2019 Paper-I
: (d) : By using Gibb’s phase rule P + F = C + 2 P + 0 = 1 + 2 P = 3
Answer
284. Gibbs phase rule for general system is :
(a) P + F = C – 1
(b) P + F = C + 1
(c) P + F = C – 2
(d) P + F = C + 2
OPSC AEE 2019 Paper-I
: (d) : Gibbs Phase rule P + F = C + 2 P = Number of Phase F = Degree of Freedom C = Number of Components
Answer
285. Which of the following options can always be approximated to be an deal gas?
(a) Highly superheated vapour
(b) Dry saturated vapour
(c) Supercritical fluid
(d) Saturated vapour
BHEL ET 2019
. (a) : Assumption of Ideal gas- • Intermolecular force should be negligible • Volume occupied by the molecular should be negligible compared to container volume. • Water vapour mixed with air is considered as ideal gas. • Steam should never be considered as ideal gas unless mentioned. • Steam is the special term given to the gases form of water only when present at high temperature.
Answer
286. Pick the correct statement about pure substances.
(a) A mixture of liquid air and gaseous air is a pure substances.
(b) A mixture of ice and liquid water is not a pure substance.
(c) A mixture of two or more phases of pure substances is not a pure substance even though the chemical composition of all the phases is the same throughout.
(d) A mixture of two or more phases of pure substance as long as the chemical composition of all the phases is the same throughout.
BHEL ET 2019
. (d) : Pure substance– A mixture of two or more phases of pure substance as long as the chemical composition of all the phases is the same throughout.
Answer
287. Water vapour can be considered as Ideal Gas.
(a) Never
(b) Always
(c) At high pressure
(d) At low pressure
OPSC AEE 2019 PAPER-II
: (d) : At very low pressure and high temperature all gases and vapour approaches to ideal gas behaviour.
Answer
288. In Van der Waals equation of state the two constant are determined from the behavior of substance at:
(a) Saturalred point
(b) Triple point
(c) Critical point
(d) Never determined
OPSC AEE 2019 PAPER - II
: (b) : In van der waals equation of state the two constant are determined from the behavior of substance at triple point. Van der waals equation ( ) 2 a P v b RT v + − =
Answer
289. Specific heat of monoatomic gases:
(a) Increase with temperature rise
(b) Decrease with temperature rise
(c) Does not depends on change In temperature
(d) None of these
OPSC AEE 2019 PAPER - II
: (b) : The molar specific heat of a gas at constant pressure (CP) is the amount of heat required to raise the temperature of 1 mol of the gas by 1ºC at the constant pressure. Its value for monoatomic ideal gas is 5 2 R and the value for diatomic ideal gas is 7R 2 50
Answer
290. A positive value of Joule-Thomson coefficient of a liquid means:
(a) Temperature drops during throttling
(b) Temperature remains constant
(c) Temperature rises during throttling
(d) None of these
OPSC AEE 2019 PAPER - II
: (a) : Joule Thomson coefficient h T P ∂ µ = ∂ 0 Temperaturerises 0 Temperatureremainsconstant 0 Temperaturedrops < µ = >
Answer
291. Joule-Thomson process is
(a) Throttling process
(b) Heating process
(c) Compression process (d) Expansion process
Gujarat PSC AE 2019
: (a) : h T constant enthalpy 'h' process. P ∂ µ = ∂
Answer
292. A gas having negative Joule Thomson coefficient (µ < 0) where throttled will
(a) become cooler
(b) become warmer
(c) remain at the same temperature
(d) either become cooler or warmer depending on the type of gas
Gujarat PSC AE 2019
: (b) : Joule Thomson coefficient h T P ∂ µ = ∂ 0 become warmer 0 remain at the same temperautre > 0 become cooler < µ = =
Answer
293. Most of the gases exhibit drop in temperature upon expansion. However, this may not be true in case of
(a) carbon dioxide
(b) oxygen
(c) hydrogen
(d) helium
BPSC AE Mains 2017 Paper - V
: (c) : In thermodynamics, the- Joule-Thomson (also known as the Joule-Kelvin effect or Kelvin-Joule effect) describes the temperature change of real gas or liquid when it is forced through a value or porous plug while keeping it insulated so that no heat is exchanged with the environment. This procedure is called a throttling process or Joule-Thomson process. At room temperature, all gases except hydrogen, helium, and neon cool upon expansion by the Joule-Thomson process when being throttled through an orifice, there three gases experience the same effect but only at lower temperature.
Answer
294. If the dryness fraction of a sample by throttling calorimeter is 0.8 and that by separating calorimeter is also 0.8, then the actual dryness fraction of sample will be taken as
(a) 0.8
(b) 0.8
(c) 0.64
(d) 0.5
BPSC AE Mains 2017 Paper - V
: (c) : x1 = Dryness fraction of steam considering separating calorimeter. x2 = Dryness fraction of steam entering the throttling calorimeter. x = x1x2 Actual dryness fraction of steam in the sample x = 0.8×0.8 = 0.64
Answer
295. Mole fraction of a component of gas mixture is equal to
(a) 1 f
(b) 2 f
(c) f
(d) f P
JPSC AE 2013 Ist Paper
. (c): The mole fraction of any component of a mixture is the ratio of the number of moles of that substance to the total number of moles of all substances present. In a mixture of gases, the partial pressure of each gas is the product of the total pressure and the mole fraction of that gas.
Answer
296. When the fuel is burned and water is release in the liquid phase, the heating value of fuel is called
(a) higher heating value
(b) lower heating value
(c) enthalpy of formation
(d) None the above
JPSC AE 2013 Ist Paper
. (a) : higher heating value
Answer
297. A fluid behave as an ideal gas provided it is at :
(a) High Temperature and Pressure
(b) High Temperature and Moderate Pressure
(c) Low Temperature and Pressure
(d) High Temperature and Low Pressure
OPSC Civil Services Pre. 2011
. (d) : A real gases obeys perfect gas law at high temperature and low pressure. Inter molecular attraction between molecules are negligible. Volume occupied by molecules as compared to total volume is negligible.
Answer
298. Match the following : 1. Empirical temperature a. Path function 2. Polarization b. Permeable to heat transfer 3. Heat transfer c. Mechanical equivalent of heat 4. Joule's experiment d. Isotherm 5. Diathermic e. Dipole moment
(a) 1–d, 2–a, 3–e, 4–c, 5–b
(b) 1–d, 2–e, 3–a, 4–c, 5–b 51
(c) 1–d, 2–b, 3–a, 4–c, 5–e
(d) 1–d, 2–c, 3–b, 4–a, 5–e
OPSC Civil Services Pre. 2011
. (b) : 1. Empirical temperature Isotherm 2. Polarization Dipole moment 3. Heat transfer Path function 4. Joule's experiment Mechanical equivalent of heat 5. Diathermic Permeable to heat transfer
Answer
299. The point that connects the saturated-liquid line to the saturated-vapour line is called the
(a) triple point
(b) critical point
(c) superheated point
(d) compressed liquid point
RPSC Vice Principal ITI 2018
. (b) : The properties of liquid and vapour are identical at critical point.
Answer
300. The enthalpy of vaporization, at critical point is
(a) maximum
(b) minimum
(c) zero
(d) none of the above
RPSC Vice Principal ITI 2018
TNPSC 2019
. (c) : The enthalpy of vaporization, at critical point is zero The line AB denotes enthalpy of vaporization. At critical point (CP), the length become zero so enthalpy of vaporization is zero. Enthalpy of vaporization hfg = hg – hf At critical point, the liquid and vapour have same properties, so hf = hg hfg= 0
Answer
301. According to Dalton's law, the total pressure of the mixture of gases is equal to
(a) greater of the partial pressure of gases
(b) average of the partial pressure of gases
(c) Sum of the partial pressure of all
(d) Sum of the partial pressure of all divided by average molecular weight
TNPSC 2019
. (c) : According to Dalton's law, the total pressure of the mixture of gases is equal to sum of the partial pressure of all.
Answer
302. Calculate the dryness fraction of steam which has 1.5 kg of water in suspension with 50 kg of steam
(a) 0.971
(b) 1
(c) 0
(d) 0.485
TNPSC 2019
. (a) : Data Given mw = 1.5 kg, mv = 50 kg Dryness fraction (x) = v w v m m m+ 50 x 0.971 1.5 50 = = +
Answer
303. Which one of the following properties remains unchanged for a real gas during JouleThomson process
(a) Temperature
(b) Enthalpy
(c) Entropy
(d) Pressure
TNPSC 2019
. (b) : Enthalpy properties remains unchanged for a real gas during Joule-Thomson process. h const. T P = ∂ = µ ∂
Answer
304. Which of the following gasses will have the maximum value of gas constant R
(a) nitrogen
(b) carbon dioxide
(c) sulpher dioxide
(d) oxygen
TSPSC AEE 2015
. (a) : Element Molecular weight (M) Nitrogen (N2) 28 Corban Dioxide (CO2) 44 Sulpher Dioxide (SO2) 64 Oxygen (O2) 32 gas constant R = R M 1 R M ∝ So, Nitrogen gas will have the maximum value of gas constant because its molecular weight is 28.
Answer
305. The specific heat of gas remains constant at all pressure and temperature. This statement pertains to
(a) Joule's law
(b) Regnault's law 52
(c) Avogadro's law
(d) Maxwell law
TSPSC AEE 2015
. (b) : According to Regnault's law, the specific heat of gas remains constant at all pressure and temperature.
Answer
306. The specific heat of an ideal gas depend on its __________alone.
(a) Pressure
(b) Volume
(c) Entropy
(d) Temperature
APPSC AEE 2016
. (d) : Specific Heat– The specific heat is the amount of heat per unit mass required to rise the temperature by one degree Celsius. ∆Q = mc∆T C = Q m T ∆ ∆ Where, m = mass (kg) C = Specific heat J/kg°C ∆Q = Change in thermal energy (J) ∆T = Change in temperature (°C)
Answer
307. The substance which is homogeneous and invariable in chemical composition throughout its mass is called as ______.
(a) ideal substance
(b) pure substance
(c) solid substance
(d) gas substance
APPSC AEE 2016
. (b) : The substance which is homogeneous and invariable in chemical composition throughout its mass is called as pure substance.
Answer
308. No liquid can exist as a liquid at
(a) 0°C temperature
(b) 200°C temperature
(c) Zero pressure
(d) Zero viscosity
Nagaland PSC CTSE 2017 Paper-2
APPSC AEE 2016
. (c) : No liquid can exist as a liquid at zero pressure:- Generally absolute zero pressure is the point where there exist a minimum temperature i.e. zero. That can be possible only when molecular momentum of system become zero. There should not be any motion of particles so there is no collision of particles, kinetic energy nullifies and the temperature becomes zero.
Answer
309. Sublimation is the process of
(a) Changing from gas state to solid state
(b) Changing from solid state to gas state
(c) Changing from liquid to vapour state
(d) Existence of solids, liquids and gases simultaneously
APPSC AEE 2016
. (b) : Sublimation– when a solid turns into a gas without first becoming liquid, that's sublimation.
Answer
310. The value of dryness fraction at critical point for water-steam phase transformation may be
(a) 0
(b) 1
(c) either 0 or 1
(d) all of these
TNPSC AE 2017
. (d) : The value of dryness fraction at critical point for water-steam phase transformation is undefined.
Answer
311. Latent heat of vaporization of water at critical point is
(a) 334 J/kg
(b) 234 J/kg
(c) 334 J/kg
(d) zero
TNPSC AE 2017
. (d) : Latent heat of vaporization of water decreases with increase in pressure and become zero at critical point [Pcr = 221 bar, Tcr = 374ºC]
Answer
312. The value of characteristics constant of oxygen would be
(a) 0.412 kJ/kg-K
(b) 0.262 kJ/kg-K
(c) 1.004 kJ/kg-K
(d) 0.624 kJ/kg-K
TNPSC AE 2018
. (b) : The value of characteristics constant of oxygen would be 0.262 kJ/kg-K.
Answer
313. It is desired to store 28 kg of nitrogen at 14 MPa pressure and 27ºC in a cylinder. Assuming that nitrogen behaves like an ideal gas, determine the size of the cylinder.
(a) 0.01782 m3
(b) 0.1782 m3
(c) 1.782 m3
(d) 17.82 m3
RPSC AE 2018
. (b) : Data given: m = 28 kg, P = 14 MPa, T = 27ºC = 300 K Ideal gas equation PV = mRT where R R M = molecular weight (M) for N2 = 28 R (Universal gas constant) = 8.314 kJ/k-m K 8.314 0.2969 / 28 R kJ K kg = = then, 28 0.2969 300 14 1000 V × × = × V = 0.178157 m3
Answer
314. The principle of working of the constant volume thermometer is based on
(a) Boyle's law
(b) Charle's law
(c) Gay – Lussac's law
(d) Equation of state
TNPSC AE 2014
. (c) : Constant volume gas thermometer– This thermometer works on the principle of Law of GayLussac. The law states that when the temperature of an ideal gas increases, there is a corresponding increase in pressure. Also, when the temperature decrease, the pressure too decrease correspondingly. 53
Answer
315. The heating of a gas at constant pressure is governed by
(a) Boyle's law
(b) Charles's law
(c) Gay-Lussac law
(d) Joule's law
TNPSC AE 2014
. (b) : Charles's Law—The heating of a gas at constant pressure. Boyle's Law—The heating of a gas at constant temperature. Gay-Lussac Law—The heating of a gas at constant volume.
Answer
316. The heating of wet steam at constant temperature till it becomes dry saturated is similar to that of heating at a
(a) constant volume
(b) constant pressure
(c) constant entropy
(d) constant enthalpy
TNPSC AE 2014
. (b) : The heating of wet steam at constant temperature till it becomes dry saturated is similar to that of heating at a constant pressure.
Answer
317. The dryness fraction of steam is equal to
(a) g g f M M + M
(b) f g f M M + M
(c) g f M M
(d) f g M M Where Mg = mass of dry steam Mf = mass of wet steam
TNPSC AE 2014
TNPSC 2019
. (a) : g g f M x M M = +
Answer
318. With the increase of pressure
(a) The boiling point of water decreases and enthalpy of evaporation increases
(b) The boiling point of water increases and enthalpy of evaporation decreases
(c) Both the boiling point of water and enthalpy of evaporation decreases
(d) Both the boiling point of water and enthalpy of evaporation increases
TNPSC AE 2014
. (b) : With the increase of pressure the boiling point of water increases and enthalpy of evaporation decreases. h1 > h2 > h3 At critical point where saturated liquid and saturated vapour line are meet, enthalpy of evaporation become zero and liquid directly flash into vapour.
Answer
319. The ratio of PV RT is
(a) Equations of state
(b) Compressibility factor
(c) Reduced properties
(d) Critical compressibility factor
TNPSC AE 2013
. (b) : Compressibility factor (Z)– Compressibility factor is also known as the compression factor or the gas deviation factor, is a correction factor which describe the deviation of a real gas from ideal gas behaviour. It is simply defined as the ratio of the moler volume of a gas to the molar volume of an ideal gas at the same temperature and pressure. PV Z RT = For ideal gas → Z = 1
Answer
320. A certain gas has Cp value of 1968 J/kgK and Cv value of 1507 J/kgK. The value of R is
(a) 0.461 KJ/kgK
(b) 1307 J/kgK
(c) 1
(d) 461 KJ/kgK
TNPSC AE 2013
. (a) : We know that- Mayer's formula is given asC C R p V − = 1968 – 1507 = R R = 461 J/kgK R = 0.461/ kJ/kgK
Answer
321. Region inside the inversion curve is represented by : (where µ is Joule - Kelvin coefficient)
(a) Cooling region, µ < 0
(b) Heating region, µ > 0
(c) Cooling region, µ > 0
(d) Heating region, µ < 0
UPRVUNL AE 2016 54
. (c) : Joule - Kelvin coefficient (µ) h T P µ ∂ = ∂ For ideal gas µ = 0
Answer
322. Ideal gas equation may be written as [where, P = absolute pressure, v = specific volume, R = characteristic gas constant, T = absolute temperature, m = mass of gas, n = number of moles of gas, Z = compressibility factor]:
(a) Pv = nRT
(b) Pv = ZRT
(c) Pv = mRT
(d) Pv = RT
UPRVUNL AE 2016
. (d) : Compressibility factor Z = Pv RT For ideal gas Z = 1 then Pv = RT
Answer
323. Generalized compressibility chart is drawn between:
(a) Reduced pressure (Pr) on y-axis and reduced temperature (Tr) on x-axis
(b) Compressibility factor (Z) on y-axis and reduced temperature (Tr) on x-axis
(c) Compressibility factor (Z) on y-axis and reduced pressure (Pr) on x-axis
(d) Compressibility factor (Z) on x-axis and reduced pressure (Pr) on y-axis
UPRVUNL AE 2016
. (c) : Generalized compressibility chart is drawn between compressibility factor (Z) on y-axis and reduced pressure (Pr) on x-axis. PVactual RT Z = actual ideal V V Z = Z = 1 - for ideal gas Z > 1 - less compressible Z < 1 - more compressible
Answer
324. Which of the following is true statement for phase diagram of pure substance?
(a) Sublimation curve, fusion curve and vaporization curve meets at critical point
(b) Extreme points of vaporization curve are triple point and critical point
(c) Extreme points of fusion curve are triple point and critical point
(d) Fusion curve for water has positive slope
UPRVUNL AE 2016
. (b) : In phase diagram of pure substance, extreme points of vaporization curve are triple point and critical points. • Fusion curve for water has negative slope.
Answer
325. Which of the following law governs the isothermal process
(a) Boyle's Law
(b) Charle's law
(c) Joule's law
(d) Gay Lussac's law
HPPSC AE 2018
. (a) : Boyle's law governs the isothermal process. T = Constant P1V1 = P2V2 = Constant Charle's Law – P = Constant 1 2 1 2 V V T T = = Constant Gay Lussac's Law – Gay Lussac's Law states that the pressure of a given mass of a gas varies directly with absolute temperature of the gas when the volume is kept constant. 1 2 1 2 P P T T = = Constant
Answer
326. Specific heat at constant pressure can be given as (Where 'ϒ' is ratio of specific heats at constant pressure and constant volume, R is a gas constant, J is the Joule constant.)
(a) ( ) R J 1 γ γ −
(b) ( ) JR γ γ −1
(c) ( ) R J R 1 γ −
(d) ( ) R J 1 γ −
HPPSC AE 2018 55
. (a) : We know that Cp – Cv = R J ................(i) and p v C .............(ii) C =γ then Cv = Cp γ Putting the value of Cv in equation (i) Cp – Cp R J = γ [ ] p 1 R C J γ− = γ ( ) ( ) p .R .R C 1 J 1 γ γ = = γ − γ −
Answer
327. For dry saturated vapour, the value of dryness fraction will be
(a) 1.0
(b) 0.75
(c) 0.5
(d) 0
RPSC LECTURER 16.01.2016
. (a) : Dryness fraction (x) Mass of vapour Mass of vapour + Mass of liquid = v L v m x m m = + at saturated vapour line mL = 0 1 0 v v m x m = = +
Answer
328. A Saturation state of pure substance (water) is a:
(a) State from which a change of phase may occur with a change in pressure or temperature.
(b) State from which a change of phase may occur with a change in pressure and temperature.
(c) State from which a change of phase may occur with a change in pressure and volume.
(d) State from which a change of phase may occur without a change in pressure and temperature.
(e) State from which a change of phase may occur without a change in pressure or temperature.
(CGPCS Polytechnic Lecturer 2017)
. (d) : Saturation State–A saturation state of pure substance (water) is a state from which a change of phase may occur without a change in pressure and temperature.
Answer
329. The reading of temperature on the Celsius scale is 60° C. What is equivalent reading of temperature on the Fahrenheit scale?
(a) 130°F
(b) 132°F
(c) 136°F
(d) 140°F
(e) 146°F
(CGPCS Polytechnic Lecturer 2017)
. (d) : We know that C 0 100 0 − − = F 32 212 32 − − C 5 = F 32 9 − C = 60°C then 60 5 = F 32 9 − = 12 F – 32 = 108 F 140 F = °
Answer
330. Super heated vapours behave:
(a) Exactly as a gas
(b) As steam
(c) As ordinary vapour
(d) Approximately as a gas
SJVN ET 2013
. (d) : Superheated vapours behave approximately as a gas.
Answer
331. RMS velocity of hydrogen gas at NTP is:
(a) 526 m/s
(b) 932 m/s
(c) 1356 m/s
(d) 1839 m/s
SJVN ET 2013
. (d) : RMS velocity of hydrogen ( ) rms 3RT V m = At NTP R = 8.314 kJ/mol.K 56 T = 273 K m = 2 × 10-3 kg/mol rms 3 3 8.31 273 V 2 10 − × × = × = 1845.15 m/s ≈ 1840 m/s
Answer
332. During melting the volume of pure substance other than water :
(a) Decreases
(b) Increases
(c) Remains constant
(d) First increases and then decreases
TRB Polytechnic Lecturer 2017
. (b) : General Substance Fusion Curve– General substance P ve T ∂ = + ∂ Liquids → Solids [Contract] Solid → Liquid [Expends] H2O Fusion Curve– H O2 P ve T ∂ = − ∂ Solid → Liquid [Contracts] Liquid → Solid [Expends]
Answer
333. The temperature of an ideal gas always deceases during
(a) Isobaric expansion
(b) Isothermal expansion
(c) Adiabatic expansion
(d) Isentropic expansion
Nagaland PSC CTSE 2017 Paper-2
. (d) : The temperature of an ideal gas always decreasing during isentropic expansion.
Answer
334. Joule-Kelvin coefficient is given by [where T = absolute temperature, P = Pressure, s = Specific entropy, h = specific enthalpy]
(a) h T s ∂ ∂
(b) h s T ∂ ∂
(c) h T P ∂ ∂
(d) T s ∂ ∂
SJVN ET 2019
. (c) : Joule-Kelvin co-efficient (µ) is slope on temperature - pressure diagram when enthalpy remains constant. h T P ∂ µ = ∂
Answer
335. On Mollier chart, slope of an isobar on h-s diagram is equal to: [where T = absolute temperature]
(a) T4
(b) T2
(c) T
(d) T3
SJVN ET 2019
. (c) : dh = Tds + vdP, For constant pressure/ isobaric process dP = 0 So, dh = Tds or p dh T ds =
Answer
336. An ideal gas having the weight of 20 N at the temperature of 27oC and pressure of 0.206 N/mm2 (abs). The gas constant will be [Consider g = 9.81 m/s2 ]
(a) 912 kJ/kg-K
(b) 0.912 kJ/kg-K
(c) 0.4251 kJ/kg-K
(d) 425.1 kJ/kg-K
SJVN ET 2019
. (c) : PV = mRT 6 0.26 20 1 R 300 10 9.81 − × = × × or R = 425.1 J/kgK = 0.4251 kJ/kgK
Answer
337. If the degree of saturation of air is zero, the air is said to be
(a) superheated air
(b) unsaturated air
(c) saturated air
(d) atmospheric air
APPSC AEE 2016
. (c) : If the degree of saturation of air is zero, the air is said to be saturated air. 57 We know that degree of saturation (DOS) = (T1 – T2) When, T1 = T2 then, DOS = 0 It means air is said to be saturated air.
Answer
338. When a real gas follows Joule Thomson expansion process, the temperature
(a) always increases
(b) always decreases
(c) remains same
(d) may increase or decrease
(e) becomes zero
CGPSC AE 2014 -II
. (d) : When a real gas follows Joule Thomson expansion process, the temperature may increase or decrease. Joule- Thomson coefficient ( µ )- Joule- Thomson Coefficient (µ ) is defined as the ratio of the temperature change to the pressure drop, and is expressed in terms of the thermal expansion coefficient. The Joule - Thomson Coefficient will be zero at a point called Inversion point for all real gases, Expansion of most real gases causes cooling when the jouleThomson coefficient is positive and the gas temperature is below the Inversion temperature. However, at atmosphere pressure, as the inversion temperature for hydrogen is low and hence hydrogen will warm during a Joule- Thomson expansion at room temperature. Since there is no change of temperature when an ideal gas expands through throttling device, a non-zero JouleThomson coefficient refers to a real gas.
Answer
339. For achieving the cooling effect by JouleKelvin, expansion the initial temperature of gas must be below the
(a) Boiling point temperature
(b) Freezing point temperature
(c) Maximum inversion temperature
(d) Saturation temperature
TNPSC AE 2013
. (c) : For achieving the cooling effect by JouleKelvin, expansion the initial temperature of gas must be below the maximum inversion temperature.
Answer
340. The kinetic energy lost in friction is transformed into heat which tends to
(a) cool or condense the steam
(b) dry or superheat the steam (c) increase the pressure of the steam
(d) reduce the dryness fraction
(e) decrease the specific volume of steam
CGPSC AE 2014 -II
. (b) : The kinetic energy lost in friction is transformed into heat which tends to dry or superheat the steam.
Answer
341. Under thermal equilibrium, flow of steam is
(a) isentropic
(b) adiabatic
(c) hyperbolic
(d) polytropic
BPSC AE 2012 Paper - V
: (a) : Under thermal equilibrium flow of steam is isentropic, because there is no heat transfer between system and surrounding.
Answer
342. Critical pressure of a liquid is the pressure
(a) above which liquid will remain liquid
(b) above which liquid will become gas
(c) above which liquid becomes vapour
(d) above which liquid becomes solid
BPSC AE 2012 Paper - V
: (c) : Critical pressure of a liquid is the pressure above which liquid becomes vapour.
Answer
343. A gas mixture consists of 3 kg of O2, 5kg of N2 and 12 kg of CH4. The mass fraction and mole fraction and mole fraction of O2 are
(a) 0.25 and 0.125
(b) 0.15 and 0.092
(c) 0.25 and 0.092
(d) 0.15 and 0.125
ESE 2019
. (b) : Mass fraction of O2 = o2 o N CH 2 2 4 m m m m + + = 3 3 5 12 + + = 0.15 Mol fraction of O2 = o2 o N CH 2 2 4 n n n n + + = 3 32 3 5 12 32 28 6 + + = 0.092
Answer
344. The ordinate and abscissa of the diagram to depict the isobaric processes of an ideal gas as a hyperbola are, respectively
(a) temperature and entropy
(b) internal energy and volume
(c) temperature and density
(d) enthalpy and entropy
ESE 2018
. (c) : For an ideal gas PV = mRT P = ρRT = constant ρT = constant i.e. equation of hyperbola.
Answer
345. Consider the following statements:
1. The entropy of a pure crystalline substance at absolute zero temperature is zero.
2. The efficiency of a reversible heat engine is independent of the nature of the working 58 substance and depends only on the temperature of the reservoirs between which it operates.
3. Carnot’s theorem states that of all heat engines operating between a given constant temperature source and a given constant temperature sink, none has a higher efficiency than a reversible engine. Which of the above statements are correct?
(a) 1 and 2 only
(b) 1 and 3 only
(c) 2 and 3 only
(d) 1, 2 and 3
ESE 2018
. (d) : (i) The third law of thermodynamics defines the absolute zero of entropy. The entropy of a pure crystalline substance at absolute zero temperature is zero. (ii) Efficiency of Carnot = L H T 1 T − Here, it is cleared that efficiency of Carnot engine depends upon the temperature of reservoirs. Working substance does not play any role in the efficiency of Carnot engine. Every heat engine works between two temperature limits and give some work out. Carnot is a ideal case of engines that is not possible. No heat engine can be efficient as Carnot.
Answer
346. Which one of the following substances has constant specific heat at all pressures and temperature?
(a) Mono-atomic gas
(b) Di-atomic gas
(c) Tri-atomic gas
(d) Poly-atomic gas
ESE 2018
. (a) : The specific heats, cp and cv vary with the temperature, the variation being different for each gas. For monoatomic gases, such as He, Ne, Ar and most metallic vapours, specific heats are constant.
Answer
347. Statement
(I): The specific heat at constant pressure for an ideal gas is always greater than the specific heat at constant volume.
Statement (II): Heat added at constant volume is not utilized for doing any external work.
ESE 2017
. (a) : When gas is heated at constant pressure (CP) it requires more heat energy as there is change is in internal energy as well as external work done. But in case of constant volume (CV) there is no external work done, so that the given heat has to increase only external energy. Therefore CP is always greater than CV. CP = P Const. dh dT = CV = V Const. du dT = We know that enthalpy of fluid is more in value as compared to internal energy of the fluid. h > u So, CP > CV
Answer
348. Statement (I): A homogeneous mixture of gases that do not react within themselves can be treated as a pure substance.
Statement (II): Flue gases can be treated as a homogeneous mixture of gases.
ESE 2017
. (a) : The composition of pure substance is invariable and same through out the sample i.e. constituents of pure substance do not react themselves. Hence statement-I is definition of pure substance and statement-II is example, so both are true.
Answer
349. At the critical point, any substance
(a) will exist in all the three phases simultaneously
(b) will change directly from solid to vapour
(c) will lose phase distinction between liquid and vapour
(d) will behave as an ideal gas
UKPSC AE 2012 Paper–II
. (c) : will lose phase distinction between liquid and vapour
Answer
350. Triple point and critical point pressure of carbon dioxide are:-
(a) 4.58mm Hg and 221.2 bar respectively
(b) 5.18 bar and 221.2 bar respectively
(c) 1 bar and 50 bar respectively
(d) 5.18 bar and 73.8 bar respectively
UKPSC AE-2013, Paper-II
. (d) : Triple point and critical point pressure of carbon dioxide are 5.18 bar and 73.8 bar respectively.
Answer
351. Which one of the following is weaker than hydrogen bonds?
(a) Ionic bond
(b) Vander Waals bond
(c) Covalent bond
(d) Metallic bond
UKPSC AE-2013, Paper-I
. (b) : Vander walls bond is weaker than hydrogen bonds.
Answer
352. The latent heat of steam with increase in pressure
(a) does not change
(b) increases
(c) decreases
(d) remains unpredictable
UKPSC AE 2007 Paper -II
. (c) : Decreases
Answer
353. An ideal gas is heated from temperature T1 and T2 by keeping its volume constant. The gas is expanded back to its initial temperature according to the law PVn = C. If the entropy change in the two processes are equal, then the value of n in terms of the adiabatic index γ is (a) 1 2 n γ + =
(b) 1 2 n γ − =
(c) 2 4 n γ + =
(d) 4 2 n γ + =
JPSC AE 2013 Ist Paper 59
. (a)
Answer
354. What is the lowest pressure at which water can exist in liquid phase in stable equilibrium?
(a) 101.325 kPa
(b) 0.311 kPa
(c) 22.06 kpa
(d) 0.611 kPa
BHEL ET 2019
. (d) : The lowest pressure at which water can exist in liquid phase in stable equilibrium of 0.611 kPa.
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