Molecular volume of any perfect gas at 600 x 103 N/m2 and 27°C will be  (a) 4.17m3/kgmol  (b) 400 m3/kg mol  (c) 0.15 m3/kg mol  (d) 41.7 m3/kg mol  (e) 417m3/kgmol.

Molecular volume of any perfect gas at 600 x 103 N/m2 and 27°C will be  (a) 4.17m3/kgmol  (b) 400 m3/kg mol  (c) 0.15 m3/kg mol  (d) 41.7 m3/kg mol  (e) 417m3/kgmol.
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A perfect gas at 27°C is heated at constant pressure till its volume is double. The final temperature is  (a) 54°C  (b) 327°C  (c) 108°C  (d) 654°C  (e) 600°C

Description : A perfect gas at 27°C is heated at constant pressure till its volume is double. The final temperature is  (a) 54°C  (b) 327°C  (c) 108°C  (d) 654°C  (e) 600°C

Last Answer : Answer : b

A 30-m vertical column of fluid (density 1878 kg/m3 ) is located where g= 9.65 mps2 . Find the pressure at the base of the column. (Formula: pg= gρhg/k )  a. 543680 N/m2  b. 543.68 kPa (gauge)  c. Both a & b  d. None of the above

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Last Answer : Both a & b

In a non-flow reversible process for which p = (- 3V+ 15) x 105N/m2,V changes from 1 m to 2 m3. The work done will be about  (a) 100 xlOO5 joules  (b) lxlO5 joules  (c) 10 xlO5 joules  (d) 10 xlO5 kilo joules  (e) 10xl04ki\ojoules.

Description : In a non-flow reversible process for which p = (- 3V+ 15) x 105N/m2,V changes from 1 m to 2 m3. The work done will be about  (a) 100 xlOO5 joules  (b) lxlO5 joules  (c) 10 xlO5 joules  (d) 10 xlO5 kilo joules  (e) 10xl04ki\ojoules.

Last Answer : Answer : c

A gas is enclosed in a cylinder with a weighted piston as the top boundary. The gas is heated and expands from a volume of 0.04 m3 to 0.10 m3 at a constant pressure of 200 kPa. Find the work done on the system.  a. 5 kJ  b. 15 kJ  c. 10 kJ  d. 12 kJ

Description : A gas is enclosed in a cylinder with a weighted piston as the top boundary. The gas is heated and expands from a volume of 0.04 m3 to 0.10 m3 at a constant pressure of 200 kPa. Find the work done on the system.  a. 5 kJ  b. 15 kJ  c. 10 kJ  d. 12 kJ

Last Answer : 12 kJ

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  (e) all systems can be regarded as closed systems.

Description : 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 ... gases have two values of specific heat  (e) all systems can be regarded as closed systems.

Last Answer : Answer : a

A vessel with a volume of cubic meter contains liquid water and water vapor ion equilibrium at 600 kPa. The liquid water has mass of1kg. Using the steam table, calculate the mass of the water vapor.  A. 0.99kg  B. 1.57 kg  C. 2.54 kg  D. 3.16 kg Formula: from the steam table at 600kPa vƒ = 0.001101 m^3 / kg vg = 0.3157 m^3 / kg Vtot = mƒ vƒ + mg vg mg = (tot-mƒ vƒ) / vg

Description : A vessel with a volume of cubic meter contains liquid water and water vapor ion equilibrium at 600 kPa. The liquid water has mass of1kg. Using the steam table, calculate the mass of the water vapor.  A. 0.99kg  B. 1.57 kg  C ... / kg vg = 0.3157 m^3 / kg Vtot = mƒ vƒ + mg vg mg = (tot-mƒ vƒ) / vg

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Utilizing the answer to the previous problem, estimate the overall or average increase in temperature ( ΔT) of the concrete roof from the energy absorbed from the sun during a12hour day. Assume that all of the radiation absorbed goes into heating the roof. The specific heat of concrete is about 900 J/kg, and the density is about 2,300 kg/m3 .  a. 7.9 °C  b. 8.9°C  c. 9.9°C  d. 10.9°C formula: ΔQ = m c ΔT

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Last Answer : 7.9 °C

A problem Drum ( 3 ft. diameter ; 6 ft. height ) is field with a fluid whose density is 50 lb/ft^3. Determine the total volume of the fluid.  A. 42.41 ft^3  B.44.35 ft^3  C.45.63 ft^3  D.41.23 ft^3 Formula: Vf = (pi d^2 h) / 4

Description : A problem Drum ( 3 ft. diameter ; 6 ft. height ) is field with a fluid whose density is 50 lb/ft^3. Determine the total volume of the fluid.  A. 42.41 ft^3  B.44.35 ft^3  C.45.63 ft^3  D.41.23 ft^3 Formula: Vf = (pi d^2 h) / 4

Last Answer : 42.41 ft^3

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  (e) unity.

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Last Answer : Answer : d

According to Avogadro’s law, for a given pressure and temperature, each molecule of a gas  (a) occupies volume proportional to its molecular weight  (b) occupies volume proportional to its specific weight  (c) occupies volume inversely proportional to its molecular weight  (d) occupies volume inversely proportional to its specific weight  (e) occupies same volume.

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Last Answer : Answer : e

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  (e) Joule’s law.

Description : 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  (e) Joule’s law.

Last Answer : Answer : a

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Last Answer : Answer : c

According to Gay-Lussac law for a perfect gas, the absolute pressure of given mass varies directly as  (a) temperature  (b) absolute  (c) absolute temperature, if volume is kept constant (d) volume, if temperature is kept constant  (e) remains constant,if volume and temperature are kept constant.

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Last Answer : Answer : c

Which of the following variables controls the physical properties of a perfect gas  (a) pressure  (b) temperature  (c) volume  (d) all of the above  (e) atomic mass.

Description : Which of the following variables controls the physical properties of a perfect gas  (a) pressure  (b) temperature  (c) volume  (d) all of the above  (e) atomic mass.

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3.0 lbm of air are contained at 25 psia and 100 ˚F. Given that Rair = 53.35 ft-lbf/lbm- ˚F, what is the volume of the container?  A.10.7 ft^3  B.14.7 ft^3  C.15 ft^3  D.24.9 ft^3 Formula: use the ideal gas law pV = mRT T = (100 +460) ˚R V = mRT/p

Description : 3.0 lbm of air are contained at 25 psia and 100 ˚F. Given that Rair = 53.35 ft-lbf/lbm- ˚F, what is the volume of the container?  A.10.7 ft^3  B.14.7 ft^3  C.15 ft^3  D.24.9 ft^3 Formula: use the ideal gas law pV = mRT T = (100 +460) ˚R V = mRT/p

Last Answer : 24.9 ft^3

For a perfect gas, according to Boyle’s law (where p = Absolute pressure, v = Volume, and T = Absolute temperature)  A. p v = constant, if T is kept constant  B. v/T = constant, if p is kept constant  C. p/T = constant, if v is kept constant  D. T/p = constant, if v is kept constant

Description : For a perfect gas, according to Boyle’s law (where p = Absolute pressure, v = Volume, and T = Absolute temperature)  A. p v = constant, if T is kept constant  B. v/T = constant, if p is kept constant  C. p/T = constant, if v is kept constant  D. T/p = constant, if v is kept constant

Last Answer : Answer: A

The absolute pressure of a given mass of a perfect gas varies inversely as its volume, when the temperature remains constant. This statement is known as Charles’ law.  A. Yes  B. No

Description : The absolute pressure of a given mass of a perfect gas varies inversely as its volume, when the temperature remains constant. This statement is known as Charles’ law.  A. Yes  B. No

Last Answer : Answer: B

According to Gay-Lussac law, the absolute pressure of a given mass of a perfect gas varies __________ as its absolute temperature, when the volume remains constant.  A. directly  B. indirectly

Description : According to Gay-Lussac law, the absolute pressure of a given mass of a perfect gas varies __________ as its absolute temperature, when the volume remains constant.  A. directly  B. indirectly

Last Answer : Answer: A

The value of 1 mm of Hg is equal to  A. 1.333 N/m2  B. 13.33 N/m2  C. 133.3 N/m2  D. 1333 N/m2

Description : The value of 1 mm of Hg is equal to  A. 1.333 N/m2  B. 13.33 N/m2  C. 133.3 N/m2  D. 1333 N/m2

Last Answer : Answer: C

A certain fluid is flowing in a 0.5m x 0.3 channel at the rate of 3 m/s and has a specific volume of 0.012 m³/kg. Determined the mass of water flowing in kg/s.  a. 267 kg/s  b. 378 kg/s  c. 375 kg/s  d. 456.5 kg/s m = Aν/V

Description : A certain fluid is flowing in a 0.5m x 0.3 channel at the rate of 3 m/s and has a specific volume of 0.012 m³/kg. Determined the mass of water flowing in kg/s.  a. 267 kg/s  b. 378 kg/s  c. 375 kg/s  d. 456.5 kg/s m = Aν/V

Last Answer : 375 kg/s

The ratio of two specific heats of air is equal to  (a) 0.17  (b) 0.24  (c) 0.1  (d) 1.41  (e) 2.71.

Description : The ratio of two specific heats of air is equal to  (a) 0.17  (b) 0.24  (c) 0.1  (d) 1.41  (e) 2.71.

Last Answer : Answer : d

Specific heat of air at constant pressure is equal to  (a) 0.17  (b) 0.21  (c) 0.24  (d) 1.0  (e) 1.41

Description : Specific heat of air at constant pressure is equal to  (a) 0.17  (b) 0.21  (c) 0.24  (d) 1.0  (e) 1.41

Last Answer : Answer : c

There are 1.36 kg of gas, for which R = 377 J/kg.k and k = 1.25, that undergo a nonflow constant volume process from p1 = 551.6 kPa and t1 = 60°C to p2 = 1655 kPa. During the process the gas is internally stirred and there are also added 105.5 kJ of heat. Determine t2. (Formula: T2= T1p2/ p1)  a. 999 K  b. 888 K  c. 456 K  d. One of the above

Description : There are 1.36 kg of gas, for which R = 377 J/kg.k and k = 1.25, that undergo a nonflow constant volume process from p1 = 551.6 kPa and t1 = 60°C to p2 = 1655 kPa. During the process the gas is internally stirred and ... (Formula: T2= T1p2/ p1)  a. 999 K  b. 888 K  c. 456 K  d. One of the above

Last Answer : 999 K

The specific heat at constant volume is  A. the amount of heat required to raise the temperature of unit mass of gas through one degree, at constant pressure  B. the amount of heat required to raise the temperature of unit mass of gas through one degree, at constant volume  C. the amount of heat required to raise the temperature of 1 kg of water through one degree  D. any one of the above

Description : The specific heat at constant volume is  A. the amount of heat required to raise the temperature of unit mass of gas through one degree, at constant pressure  B. the amount of heat required to raise ... to raise the temperature of 1 kg of water through one degree  D. any one of the above

Last Answer : Answer: B

To convert volumetric analysis to gravimetric analysis, the relative volume of each constituent of the flue gases is  (a) divided by its molecular weight  (b) multiplied by its molecular weight  (c) multiplied by its density  (d) multiplied by its specific weight  (e) divided by its specific weight.

Description : To convert volumetric analysis to gravimetric analysis, the relative volume of each constituent of the flue gases is  (a) divided by its molecular weight  (b) multiplied by its molecular weight  (c) ... by its density  (d) multiplied by its specific weight  (e) divided by its specific weight.

Last Answer : Answer : b

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Description : According to which law, all perfect gases change in volume by l/273th of their original volume at 0°C for every 1°C change in temperature when pressure remains constant  (a) Joule’s law  (b) Boyle’s law  (c) Regnault’s law  (d) Gay-Lussac law  (e) Charles’ law.

Last Answer : Answer : e

A cylinder contains oxygen at a pressure of 10 atm and a temperature of 300 K. The volume of the cylinder is 10 liters. What is the mass of the oxygen in grams? Molecular weight (MW) of oxygen is 32 g/mole?  a. 125.02  b. 130.08  c. 135.05  d. 120.04

Description : A cylinder contains oxygen at a pressure of 10 atm and a temperature of 300 K. The volume of the cylinder is 10 liters. What is the mass of the oxygen in grams? Molecular weight (MW) of oxygen is 32 g/mole?  a. 125.02  b. 130.08  c. 135.05  d. 120.04

Last Answer : 130.08 {(10atm)(10)(32)/(0.0821) (300K)}

The pressure gauge on a 2000 m³ tank of oxygen gas reads 600 kPa. How much volumes will the oxygen occupied at pressure of the outside air 100 kPa?  a) 14026.5 m³  b) 15026.5 m³  c) 13026.5 m³  d) 16026.5 m³ Formula: P1V1/T1 =P2V2/T2

Description : The pressure gauge on a 2000 m³ tank of oxygen gas reads 600 kPa. How much volumes will the oxygen occupied at pressure of the outside air 100 kPa?  a) 14026.5 m³  b) 15026.5 m³  c) 13026.5 m³  d) 16026.5 m³ Formula: P1V1/T1 =P2V2/T2

Last Answer : 14026.5 m³

One kg of carbon requires 4/3 kg of oxygen and produces __________ kg of carbon monoxide gas.  A. 8/3  B. 11/3  C. 11/7  D. 7/3

Description : One kg of carbon requires 4/3 kg of oxygen and produces __________ kg of carbon monoxide gas.  A. 8/3  B. 11/3  C. 11/7  D. 7/3

Last Answer : Answer: D

One kg of carbon monoxide requires __________ kg of oxygen to produce 11/7 kg of carbon dioxide gas.  A. All  B. 4/7  C. 11/4  D. 9/7

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Last Answer : Answer: B

What is the specific weight of water at standard condition? (Formula: γ = ρg / k)  a. 1000 kgm/m3  b. 9.8066 m/s2  c. 1000 kgf/m3  d. None of the above

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Last Answer : 1000 kgf/m3

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  (e) 735 J/kmol °K.

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Last Answer : Answer : b

The value of the product of molecular weight and the gas characteristic constant for all the gases in M.K.S. unit is  (a) 29.27 kgfm/mol°K  (b) 8314kgfm/mol°K  (c) 848kgfm/mol°K  (d) 427kgfm/mol°K  (e) 735 kgfm/mol°K.

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Last Answer : Answer : c

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Last Answer : Answer : d

Which of the following laws is applicable for the behavior of a perfect gas  (a) Boyle’s law  (b) Charles’law  (c) Gay-Lussac law  (d) all of the above  (e) Joule’s law.

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Last Answer : Answer : d

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Last Answer : 15.7 BTU

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Last Answer : Answer: C

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Last Answer : 1.38

A 1-kg steam-water mixture at 1.0 MPa is contained in an inflexible tank. Heat is added until the pressure rises to 3.5 MPa and the temperature to 400°. Determine the heat added.  a) 1378.7 kJ  b) 1348.5 kJ  c) 1278,7 kJ  d) 1246,5 kJ Formula: Q = (h2 – p2v2) –(h1 –p1v1)

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Last Answer : 1378.7 kJ

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Description : The molecular number density of an ideal gas at standard temperature and pressure in cm3  a. Froude number  b. Loschmidt number  c. Mach number  d. Reynold number

Last Answer : Loschmidt number

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Description : The energy associated with individual molecules in a gas, liquid or solid.  a. Specific Energy  b. Molecular Energy  c. Internal Energy  d. Phase Energy

Last Answer : Internal Energy

Which of the following gas has a minimum molecular mass?  A. Oxygen  B. Nitrogen  C. Hydrogen  D. Methane

Description : Which of the following gas has a minimum molecular mass?  A. Oxygen  B. Nitrogen  C. Hydrogen  D. Methane

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According to Avogadro’s law  A. the product of the gas constant and the molecular mass of an ideal gas is constant  B. the sum of partial pressure of the mixture of two gases is sum of the two  C. equal volumes of all gases, at the same temperature and pressure, contain equal number of molecules  D. all of the above

Description : According to Avogadro's law  A. the product of the gas constant and the molecular mass of an ideal gas is constant  B. the sum of partial pressure of the mixture of two gases is sum of the ... all gases, at the same temperature and pressure, contain equal number of molecules  D. all of the above

Last Answer : Answer: C

The universal gas constant of a gas is the product of molecular mass of the gas and the gas constant.  A. Correct  B. Incorrect

Description : The universal gas constant of a gas is the product of molecular mass of the gas and the gas constant.  A. Correct  B. Incorrect

Last Answer : Answer: A

The SI unit of mass  a. kg  b. g  c. N  d. None of the above

Description : The SI unit of mass  a. kg  b. g  c. N  d. None of the above

Last Answer : kg

Two masses, one of the 10kg and the other unknown, are placed on a scale in a region where g = 9.67 m/sec2 . The combined weight of these two masses is 174.06 N. Find the unknown mass in kg.  a. 20 kg  b. 19 kg  c. 18 kg  d. 17 kg formula: m=Fg k / g

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A perfect gas has a value of R= 319.2 J/ kf.K and k= 1.26. If 120 kJ are added to 2.27 kf\g of this gas at constant pressure when the initial temp is 32.2°C? Find T2.  a. 339.4 K  b. 449.4 K  c. 559.4K  d. 669.4K formula: cp = kR/ k-1 Q= mcp(T2-T1)

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Last Answer : 339.4 K

An ideal gas whose specific heats are constant is called _____.  A. Perfect gas  B. Natural gas  C. Artificial gas  D. Refined gas

Description : An ideal gas whose specific heats are constant is called _____.  A. Perfect gas  B. Natural gas  C. Artificial gas  D. Refined gas

Last Answer : Perfect gas

The behaviour of a perfect gas, undergoing any change in the variables which control physical properties, is governed by  A. Boyle’s law  B. Charles’ law  C. Gay-Lussac law  D. all of these

Description : The behaviour of a perfect gas, undergoing any change in the variables which control physical properties, is governed by  A. Boyle’s law  B. Charles’ law  C. Gay-Lussac law  D. all of these

Last Answer : Answer: D

According to Gay-Lussac law for a perfect gas, p/T = constant, if v is kept constant.  A. True  B. False

Description : According to Gay-Lussac law for a perfect gas, p/T = constant, if v is kept constant.  A. True  B. False

Last Answer : Answer: A