On weight basis, air contains following parts of oxygen  (a) 21  (b) 23  (c) 25  (d) 73  (e) 79.

On weight basis, air contains following parts of oxygen
 (a) 21
 (b) 23
 (c) 25
 (d) 73
 (e) 79.
by

1 Answer

Answer :

Answer : b
by

Related questions

On volume basis, air contains following parts of oxygen  (a) 21  (b) 23  (c) 25  (d) 77  (e) 79.

Description : On volume basis, air contains following parts of oxygen  (a) 21  (b) 23  (c) 25  (d) 77  (e) 79.

Last Answer : Answer : a

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 volume of the gas held at constant pressure increases 4 cm² at 0°C to 5cm². What is the final pressure?  a. 68.65ºC  b. 68.25ºC  c. 70.01°C  d. 79.1ºC t2= T2–T1

Description : The volume of the gas held at constant pressure increases 4 cm² at 0°C to 5cm². What is the final pressure?  a. 68.65ºC  b. 68.25ºC  c. 70.01°C  d. 79.1ºC t2= T2–T1

Last Answer : 981 N

From the steam table, determine the average constant pressure specific heat (c) of steam at 10 kPa and45.8 ˚C  A.1.79 kJ/ kg-˚C  B.10.28 kJ/ kg-˚C  C.30.57 kJ/ kg-˚C  D. 100.1 kJ/ kg-˚C Formula: h = c T ∆ ∆ From the steam table At 47.7 ˚C h= 2588.1 kJ/ kg At 43.8 ˚C h= 2581.1 kJ/ kg

Description : From the steam table, determine the average constant pressure specific heat (c) of steam at 10 kPa and45.8 ˚C  A.1.79 kJ/ kg-˚C  B.10.28 kJ/ kg-˚C  C.30.57 kJ/ kg-˚C  D. 100.1 kJ/ kg-˚C Formula: h = c T ∆ ∆ From the steam table At 47.7 ˚C h= 2588.1 kJ/ kg At 43.8 ˚C h= 2581.1 kJ/ kg

Last Answer : 1.79 kJ/ kg-˚C

Find the pressure due to a column of mercury 74.0 cm high.  a. 91.80 x 10^3 N/m^2  b. 73.56 x 10^2 N/m^2  c. 9.86 x 10^4 N/m^2  d. 87.2 x 10^4 N/m^2

Description : Find the pressure due to a column of mercury 74.0 cm high.  a. 91.80 x 10^3 N/m^2  b. 73.56 x 10^2 N/m^2  c. 9.86 x 10^4 N/m^2  d. 87.2 x 10^4 N/m^2

Last Answer : 9.86 x 10^4 N/m^2

What is the value of the universal gas constant in kJ/kmol K? ∙  A. 10.73  B. 1.986  C. 8.314  D. 1545

Description : What is the value of the universal gas constant in kJ/kmol K? ∙  A. 10.73  B. 1.986  C. 8.314  D. 1545

Last Answer : 8.314

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

An ideal gas at 45psig and 80ºF is heated in the close container to 130ºF. What is the final pressure?  a. 65.10 psi  b. 65.11 psi  c. 65.23 psi  d. 61.16 psi P1V1/T1= P2V2/T2;V = Constant

Description : An ideal gas at 45psig and 80ºF is heated in the close container to 130ºF. What is the final pressure?  a. 65.10 psi  b. 65.11 psi  c. 65.23 psi  d. 61.16 psi P1V1/T1= P2V2/T2;V = Constant

Last Answer : 65.23 psi

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

Refer to problem # 11. Determine the force that accelerates if to 12 m/s^2. horizontally along frictionless plane.  A. 2474.23 N  B. 2574.23 N  C. 3474.23 N  D. 2374.23 N Formula : M = wk / g F = ma /k

Description : Refer to problem # 11. Determine the force that accelerates if to 12 m/s^2. horizontally along frictionless plane.  A. 2474.23 N  B. 2574.23 N  C. 3474.23 N  D. 2374.23 N Formula : M = wk / g F = ma /k

Last Answer : 2474.23 N

A system weighing 2kN. Determine the force that accelerate if to 12 m/s^2. a. vertically upward when g = 9.7 m/s^2  A. 4474.23 N  B.5484.23 N  C.4495.23 N  D.5488.23 N Formula: F = m/k (a +g)

Description : A system weighing 2kN. Determine the force that accelerate if to 12 m/s^2. a. vertically upward when g = 9.7 m/s^2  A. 4474.23 N  B.5484.23 N  C.4495.23 N  D.5488.23 N Formula: F = m/k (a +g)

Last Answer : 4474.23 N

Which one is the correct relation between energy efficiency ratio (EER) and coefficient of performance (COP)?  A. EER = 2.34 COP  B. EER = 3.24 COP  C. EER = 3.42 COP  D. EER = 4.23 COP

Description : Which one is the correct relation between energy efficiency ratio (EER) and coefficient of performance (COP)?  A. EER = 2.34 COP  B. EER = 3.24 COP  C. EER = 3.42 COP  D. EER = 4.23 COP

Last Answer : EER = 3.42 COP

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

The basis for measuring thermodynamic property of temperature is given by  (a) zeroth law of thermodynamics  (b) first law of thermodynamics  (c) second law of thermodynamics  (d) third law of thermodynamics  (e) Avogadro’s hypothesis.

Description : The basis for measuring thermodynamic property of temperature is given by  (a) zeroth law of thermodynamics  (b) first law of thermodynamics  (c) second law of thermodynamics  (d) third law of thermodynamics  (e) Avogadro’s hypothesis.

Last Answer : Answer : a

Twenty grams of ice at 0˚C melts to water at 0˚C. How much does the entropy of the 20g change in this process?  a. 30.5 J/K  b. 24.6 J/K  c. 21.3 J/K  d. 15.7 J/K

Description : Twenty grams of ice at 0˚C melts to water at 0˚C. How much does the entropy of the 20g change in this process?  a. 30.5 J/K  b. 24.6 J/K  c. 21.3 J/K  d. 15.7 J/K

Last Answer : 24.6 J/K

Two 12 ft. sections of aluminum siding are placed end to end on the outside wall of house. How large a gap should be left between the pieces to prevent buckling if the temperature can change by 55˚C?  a. 0.21 m  b. 0.18 m  c. 0.31 in  d. 0.18 in

Description : Two 12 ft. sections of aluminum siding are placed end to end on the outside wall of house. How large a gap should be left between the pieces to prevent buckling if the temperature can change by 55˚C?  a. 0.21 m  b. 0.18 m  c. 0.31 in  d. 0.18 in

Last Answer : 0.18 in

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

The Law of Thermodynamics that provides the basis for measuring the thermodynamic property of temperature.  a. Charle’s Law  b. Boyle’s Law  c. Zeroth Law  d. Gas Law

Description : The Law of Thermodynamics that provides the basis for measuring the thermodynamic property of temperature.  a. Charle’s Law  b. Boyle’s Law  c. Zeroth Law  d. Gas Law

Last Answer : Zeroth Law

Roughly what is the total weight of air in the entire earth?  a. 1 x 10^5 tons  b. 2 x 10^6 tons  c. 6 x 10^15 tons  d. 8 x 10^10 tons

Description : Roughly what is the total weight of air in the entire earth?  a. 1 x 10^5 tons  b. 2 x 10^6 tons  c. 6 x 10^15 tons  d. 8 x 10^10 tons

Last Answer : 6 x 10^15 tons

Let a closed system execute a state change for which the heat is Q = 100 J and work is W = -25 J. Find E. ∆ (Formula: E = Q- W) ∆  a. 125 J  b. 123 J  c. 126 J  d. None of the above

Description : Let a closed system execute a state change for which the heat is Q = 100 J and work is W = -25 J. Find E. ∆ (Formula: E = Q- W) ∆  a. 125 J  b. 123 J  c. 126 J  d. None of the above

Last Answer : 125 J

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

Last Answer : 3.16 kg

A 10m^3 vessel initially contains 5 m^3 of liquid water and 5 m^3 of saturated water vapor at 100 kPa. Calculate the internal energy of the system using the steam table.  A. 5 x10^5 kJ  B. 8x10^5 kJ  C. 1 x10^6 kJ  D. 2 x10^6 kJ Formula: fromthe steamtable vƒ = 0.001043 m^3 / kg vg = 1.6940 m^3 / kg u ƒ= 417.3 kJ/kg ug= 2506kJ/kg formula: Mvap = V vap/vg M liq = Vliq/ vƒ u =uƒM liq + ug M vap

Description : A 10m^3 vessel initially contains 5 m^3 of liquid water and 5 m^3 of saturated water vapor at 100 kPa. Calculate the internal energy of the system using the steam table.  A. 5 x10^5 kJ  B. 8x10^5 kJ  C. 1 ... 3 kJ/kg ug= 2506kJ/kg formula: Mvap = V vap/vg M liq = Vliq/ vƒ u =uƒM liq + ug M vap

Last Answer : 2 x10^6 kJ

Is the amount of a substance that contains Avogadro’s number of atoms/molecules.  a. mass  b. matter  c. gram-mole  d. volume

Description : Is the amount of a substance that contains Avogadro’s number of atoms/molecules.  a. mass  b. matter  c. gram-mole  d. volume

Last Answer : gram-mole

The thermodynamic property of a system is said to be an intensive property whose value for the entire system __________ the sum of their value for the individual parts of the system.  A. is equal to  B. is not equal to

Description : The thermodynamic property of a system is said to be an intensive property whose value for the entire system __________ the sum of their value for the individual parts of the system.  A. is equal to  B. is not equal to

Last Answer : Answer: B

Oxygen at 15ºC and 10.3 Mpa gauge pressure occupies 600L. What is the occupied by the oxygen at 8.28 Mpa gauge pressure and 35ºC?  a. 789.32 L  b. 796.32 L  c. 699 L  d. 588 L V2= P1V1/T1P2

Description : Oxygen at 15ºC and 10.3 Mpa gauge pressure occupies 600L. What is the occupied by the oxygen at 8.28 Mpa gauge pressure and 35ºC?  a. 789.32 L  b. 796.32 L  c. 699 L  d. 588 L V2= P1V1/T1P2

Last Answer : 796.32 L

Twenty grams of oxygen gas are compressed at a constant temperature of 30 ˚C to 5%of their original volume. What work is done on the system.  A.824 cal  B.924 cal  C.944 cal  D.1124 cal Formula: W = -mRTln (V2/V1) Where R = (1.98 cal/gmole·K) (32 g/gmole)

Description : Twenty grams of oxygen gas are compressed at a constant temperature of 30 ˚C to 5%of their original volume. What work is done on the system.  A.824 cal  B.924 cal  C.944 cal  D.1124 cal Formula: W = -mRTln (V2/V1) Where R = (1.98 cal/gmole·K) (32 g/gmole)

Last Answer : 1124 cal

Find the change in internal energy of 5 lb. of oxygen gas when the temperature changes from 100 ˚F to 120 ˚F. CV = 0.157 BTU/lbm-˚R  A.14.7 BTU  B.15.7 BTU  C. 16.8 BTU  D. 15.9 BTU Formula: U= mcv T

Description : Find the change in internal energy of 5 lb. of oxygen gas when the temperature changes from 100 ˚F to 120 ˚F. CV = 0.157 BTU/lbm-˚R  A.14.7 BTU  B.15.7 BTU  C. 16.8 BTU  D. 15.9 BTU Formula: U= mcv T

Last Answer : 15.7 BTU

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

Last Answer : Answer: C

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

Description : 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

Last Answer : Answer: B

The atomic mass of oxygen is  A. 12  B. 14  C. 16  D. 32

Description : The atomic mass of oxygen is  A. 12  B. 14  C. 16  D. 32

Last Answer : Answer: C

Which of the following has the minimum atomic mass?  A.Oxygen  B.Sulphur  C.Nitrogen  D.Carbon

Description : Which of the following has the minimum atomic mass?  A.Oxygen  B.Sulphur  C.Nitrogen  D.Carbon

Last Answer : Answer: D

A vertical column of water will be supported to what height by standard atmospheric pressure.  a. 33.9 ft  b. 45 ft  c. 67 ft  d. 25.46 ft ho= Po/Yo

Description : A vertical column of water will be supported to what height by standard atmospheric pressure.  a. 33.9 ft  b. 45 ft  c. 67 ft  d. 25.46 ft ho= Po/Yo

Last Answer : 33.9 ft

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

Last Answer : 443 kPa

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

What is the absorptivity of a black body?  A. 0  B. 1  C. 0.5  D. 0.25

Description : What is the absorptivity of a black body?  A. 0  B. 1  C. 0.5  D. 0.25

Last Answer : 1

What is the emissivity of a black body?  A. 0  B. 1  C. 0.5  D. 0.25

Description : What is the emissivity of a black body?  A. 0  B. 1  C. 0.5  D. 0.25

Last Answer : 1

If the count of Echinochloa in weedy check is 70 and that in a treatment 25/m2, weed control efficiency will be a). 54.3% b). 64.3% c). 73.4% d). 79.6%

Description : If the count of Echinochloa in weedy check is 70 and that in a treatment 25/m2, weed control efficiency will be a). 54.3% b). 64.3% c). 73.4% d). 79.6%

Last Answer : b). 64.3%

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.

Description : 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.

Last Answer : Answer : b

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.

Description : 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.

Last Answer : Answer : d

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.

Description : 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.

Last Answer : Answer : c

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.

Description : 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 ...  (d) occupies volume inversely proportional to its specific weight  (e) occupies same volume.

Last Answer : Answer : e

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

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

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.

Description : 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 ... all  (d) sum of the partial pressures of all divided by average molecular weight  (e) atmospheric pressure.

Last Answer : Answer : c

The following cycle is used for air craft refrigeration  (a) Brayton cycle  (b) Joule cycle  (c) Carnot cycle  (d) Bell-Coleman cycle  (e) Reversed-Brayton cycle.

Description : The following cycle is used for air craft refrigeration  (a) Brayton cycle  (b) Joule cycle  (c) Carnot cycle  (d) Bell-Coleman cycle  (e) Reversed-Brayton cycle.

Last Answer : Answer : e

Compressed air coming out from a punctured football  (a) becomes hotter  (b) becomes cooler1  (c) remains at the same temperature  (d) may become hotter or cooler depend-ing upon the humidity of the surround¬ing air  (e) attains atmospheric temperature.

Description : Compressed air coming out from a punctured football  (a) becomes hotter  (b) becomes cooler1  (c) remains at the same temperature  (d) may become hotter or cooler depend-ing upon the humidity of the surround¬ing air  (e) attains atmospheric temperature.

Last Answer : (b) becomes cooler1

The door of a running refrigerator inside a room was left open. Which of the following statements is correct?  (a) The room will be cooled to the temperature inside the refrigerator.  (b) The room will be cooled very slightly.  (c) The room will be gradually warmed up.  (d) The temperature of the air in room will remain unaffected.  (e) any one of above is possible depending on the capacity.

Description : The door of a running refrigerator inside a room was left open. Which of the following statements is correct?  (a) The room will be cooled to the temperature inside the refrigerator.  (b) The room ... air in room will remain unaffected.  (e) any one of above is possible depending on the capacity.

Last Answer : Answer : c

The specific heat of air increases with increase in  (a) temperature  (b) pressure  (c) both pressure and temperature  (d) variation of its constituents  (e) air flow

Description : The specific heat of air increases with increase in  (a) temperature  (b) pressure  (c) both pressure and temperature  (d) variation of its constituents  (e) air flow

Last Answer : Answer : a

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