The standard reference atmospheric pressure  a. 760 mmHg  b. 1 atm  c. 14.696 psia  d. All of the above

The standard reference
atmospheric pressure
 a. 760 mmHg
 b. 1 atm
 c. 14.696 psia
 d. All of the above
by

1 Answer

Answer :

All of the above
by

Related questions

14.696 psia = _____ mmHg  a. 760  b. 1  c. 350  d. None of the above

Description : 14.696 psia = _____ mmHg  a. 760  b. 1  c. 350  d. None of the above

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Which of the following is NOT a value of the standard atmospheric pressure?  A. 1 bar  B. 1 atm  C. 1 kgf/cm^2  D. 14.223 psi

Description : Which of the following is NOT a value of the standard atmospheric pressure?  A. 1 bar  B. 1 atm  C. 1 kgf/cm^2  D. 14.223 psi

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A pressure gage registers 50 psig in a region where the barometer is 14.25 psia. Find absolute pressure in psia, Pa. (Formula; p = patm+ pg)  a. 433 kPa  b. 443 kPa  c. 343 kPa  d. None of the above

Description : A pressure gage registers 50 psig in a region where the barometer is 14.25 psia. Find absolute pressure in psia, Pa. (Formula; p = patm+ pg)  a. 433 kPa  b. 443 kPa  c. 343 kPa  d. None of the above

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Helium ( R= 0.4698 BTU/lbm-˚R ) is compressed isothermally from 14.7 psia and 68 ˚F. The compression ratio is 1:4. Calculate the work done by the gas.  A. –1454 BTU/lbm  B. -364 BTU/lbm  C.-187BTU/lbm  D.46.7 BTU/lbm Formula: W = RT ln (V2/V1)

Description : Helium ( R= 0.4698 BTU/lbm-˚R ) is compressed isothermally from 14.7 psia and 68 ˚F. The compression ratio is 1:4. Calculate the work done by the gas.  A. –1454 BTU/lbm  B. -364 BTU/lbm  C.-187BTU/lbm  D.46.7 BTU/lbm Formula: W = RT ln (V2/V1)

Last Answer : -364 BTU/lbm

What horse power is required to isothermally compress 800 ft^3 of Air per minute from 14.7 psia to 120 psia?  A. 28 hp  B.108 hp  C.256 hp  D.13900 hp Formula: W= p1V1 ln (p1/p2) Power = dW / dt

Description : What horse power is required to isothermally compress 800 ft^3 of Air per minute from 14.7 psia to 120 psia?  A. 28 hp  B.108 hp  C.256 hp  D.13900 hp Formula: W= p1V1 ln (p1/p2) Power = dW / dt

Last Answer : 108 hp

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

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A vertical column of water will be supported to what height by standard atmospheric pressure. If the Y w = 62.4lb/ft3 po = 14.7 psi.  a. 44.9 ft  b. 33.9 ft  c. 22.9 ft  d. 55.9 ft formula: ho= po/Yw

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Last Answer : 33.9 ft

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Last Answer : T2=1.620ºR, ∫H = 122.83 Btu

A certain gas, with cp = 0.529Btu/lb.°R and R = 96.2 ft.lb/lb.°R, expands from 5 cu ft and 80°F to 15 cu ft while the pressure remains constant at 15.5 psia. Compute for T2. (Formula: T2= T1V2/V1)  a. 460°R  b. 270°R  c. 1620 °R  d. None of the above

Description : A certain gas, with cp = 0.529Btu/lb.°R and R = 96.2 ft.lb/lb.°R, expands from 5 cu ft and 80°F to 15 cu ft while the pressure remains constant at 15.5 psia. Compute for T2. (Formula: T2= T1V2/V1)  a. 460°R  b. 270°R  c. 1620 °R  d. None of the above

Last Answer : 1620 °R

Ten cu ft. of air at 300 psia 400°F is cooled to 140°F at constant volume. What is the final pressure? (formula: p2 = p1T2/T1)  a. 0  b. 209 psia  c. - 420 psia  d. None of the above

Description : Ten cu ft. of air at 300 psia 400°F is cooled to 140°F at constant volume. What is the final pressure? (formula: p2 = p1T2/T1)  a. 0  b. 209 psia  c. - 420 psia  d. None of the above

Last Answer : 209 psia

What is the resulting pressure when one pound of air at 15 psia and 200 ˚F is heated at constant volume to 800 ˚F?  A.15 psia  B. 28.6 psia  C. 36.4 psia.  D. 52.1 psia Formula : T1/p1 = T2/p2 p2= p1T2 / T1

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Last Answer : 28.6 psia

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Last Answer : 1.1577 BTU/lbm-˚R

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Last Answer : 190 in³

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Last Answer : 33.9 ft

_________ is defined as that equivalent to the pressure due to a column of mercury 76cm long.  a. surface pressure  b. gage pressure  c. standard atmospheric pressure  d. isobaric pressure

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Last Answer : 62.4 lb

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Description : I kgf/cm2 is equal to  (a) 760 mm Hg  (b) zero mm Hg  (c) 735.6 mm Hg  (d) 1 mm Hg  (e) l00 mm Hg.

Last Answer : Answer : c

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

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

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Description : What is the latent heat of vaporization of water at 1 atm?  A. 2314.8 kJ/kg  B. 2257.1 kJ/kg  C. 2511.7 kJ/kg  D. 2429.8 kJ/kg

Last Answer : 2257.1 kJ/kg

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Last Answer : 333.7 kJ/kg

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While swimming at depth of120m in a fresh water lake, A fish emits an air bubbles of volume 2.0mm³ atmospheric pressure is 100kPa. What is the pressure of the bubble?  a) 217.7 kPa  b) 317.7 kPa  c) 417.7 kPa  d) 517.7 kPa Formula: P= δh

Description : While swimming at depth of120m in a fresh water lake, A fish emits an air bubbles of volume 2.0mm³ atmospheric pressure is 100kPa. What is the pressure of the bubble?  a) 217.7 kPa  b) 317.7 kPa  c) 417.7 kPa  d) 517.7 kPa Formula: P= δh

Last Answer : 217.7 kPa

While swimming a depth of 13m in a fresh water lake a fish emits an air bubble of volume 2 mm² atmospheric pressure is 100kpa what is the original pressure of the bubble.  a. 217.17 kpa  b. 119 kpa  c. 326.15 kpa  d. 210 kap Pabs = Pg + Patm

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Last Answer : 217.17 kpa

What type of pressure cannot be used for Boyle’s Law?  a. Atmospheric Pressure  b. Gauge Pressure  c. Surface Pressure  d. Isobaric Pressure

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Water boils when:  A. Its saturated vapor pressure equals to the atmospheric pressure  B. Its vapor pressure equals 76cm of mercury  C. Its temperature reaches 212 degree Celsius  D. Its vapor pressure is 1 gram per sq. cm

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Last Answer : Its saturated vapor pressure equals to the atmospheric pressure

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Last Answer : Barometer

What is the pressure below atmospheric pressure called?  A. Gage pressure  B. Absolute pressure  C. Atmospheric pressure  D. Vacuum pressure

Description : What is the pressure below atmospheric pressure called?  A. Gage pressure  B. Absolute pressure  C. Atmospheric pressure  D. Vacuum pressure

Last Answer : Vacuum pressure

What is considered as the actual pressure at a given position and is measured relative to absolute vacuum?  A. Gage pressure  B. Absolute pressure  C. Atmospheric pressure  D. Vacuum pressure

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Which of the following is correct?  A. Absolute pressure = Gauge pressure + Atmospheric pressure  B. Gauge pressure = Absolute pressure + Atmospheric pressure  C. Atmospheric pressure = Absolute pressure + Gauge pressure  D. Absolute pressure = Gauge pressure – Atmospheric pressure

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Last Answer : Answer: A

One barometric pressure or 1 atmospheric pressure is equal to  (a) 1 kgf/cnr2  (b) 1.033 kgf/cm2  (c) 0 kgf/cm2  (d) 1.0197 kgf/cm2  (e) 100 kgf/cm2.

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Last Answer : Answer : b

Water contained in a beaker can be made to boil by passing steam through it  (a) at atmospheric pressure  (b) at a pressure below the firuosphejric pressure  (c) at a pressure greater than atmospheric pressure  (d) any pressure  (e) not possible.

Description : Water contained in a beaker can be made to boil by passing steam through it  (a) at atmospheric pressure  (b) at a pressure below the firuosphejric pressure  (c) at a pressure greater than atmospheric pressure  (d) any pressure  (e) not possible.

Last Answer : Answer : c

For which of the following substances, the gas laws can be used with minimum error  (a) dry steam  (b) wet steam  (c) saturated steam  (d) superheated steam  (e) steam at atmospheric pressure.

Description : For which of the following substances, the gas laws can be used with minimum error  (a) dry steam  (b) wet steam  (c) saturated steam  (d) superheated steam  (e) steam at atmospheric pressure.

Last Answer : Answer : d

According to Dalton’s law, the total pres sure of the mixture of gases is equal to  (a) greater of the partial pressures of all  (b) average of the partial pressures of all  (c) sum of the partial pressures of all  (d) sum of the partial pressures of all divided by average molecular weight  (e) atmospheric pressure.

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Last Answer : Answer : c

Gas is enclosed in a cylinder with a weighted piston as the stop boundary. The gas is heated and expands from a volume of 0.04 m^3 to 0.10 m^3 at a constant pressure of 200kPa.Calculate the work done by the system.  A. 8 kJ  B. 10 kJ  C.12 kJ  D.14 kJ Formula: W = p(V2-V1)

Description : Gas is enclosed in a cylinder with a weighted piston as the stop boundary. The gas is heated and expands from a volume of 0.04 m^3 to 0.10 m^3 at a constant pressure of 200kPa.Calculate the work done by the system.  A. 8 kJ  B. 10 kJ  C.12 kJ  D.14 kJ Formula: W = p(V2-V1)

Last Answer : 12 kJ

If air is at pressure, p, of 3200 lbf/ft2 , and at a temperature, T, of 800 ˚R, what is the specific volume, v? (R=5303 ft-lbf/lbm-˚R, and air can be modeled as an ideal gas.)  A.9.8 ft^3/lbm  B.11.2 ft^3/lbm  C.13.33 ft^3/lbm  D.14.2 ft^3/lbm Formula: pv = RT v = RT / p

Description : If air is at pressure, p, of 3200 lbf/ft2 , and at a temperature, T, of 800 ˚R, what is the specific volume, v? (R=5303 ft-lbf/lbm-˚R, and air can be modeled as an ideal gas.)  A.9.8 ft^3/lbm  B.11.2 ft^3/lbm  C.13.33 ft^3/lbm  D.14.2 ft^3/lbm Formula: pv = RT v = RT / p

Last Answer : 13.33 ft^3/lbm

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

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Last Answer : Loschmidt number

At steam point, the temperature of water and its vapor at standard pressure are:  a. Extremes or maximum  b. unity  c. in equilibrium  d. undefined

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Last Answer : in equilibrium

Which of the following is standard temperature and pressure (STP)?  A. 0 degree Celsius and one atmosphere  B. 32 degree Fahrenheit and zero pressure  C. 0 degree Kelvin and one atmosphere  D. 0 degree Fahrenheit and zero pressure

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Last Answer : 0 degree Celsius and one atmosphere

What refers to the heating of the earth’s atmosphere not caused by direct sunlight but by infrared light radiated by the surface and absorbed mainly by atmospheric carbon dioxide?  A. Greenhouse effect  B. Global warming  C. Thermal rise effect  D. Ozone effect

Description : What refers to the heating of the earth’s atmosphere not caused by direct sunlight but by infrared light radiated by the surface and absorbed mainly by atmospheric carbon dioxide?  A. Greenhouse effect  B. Global warming  C. Thermal rise effect  D. Ozone effect

Last Answer : Greenhouse effect