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

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
by

1 Answer

Answer :

209 psia
by

Related questions

Ten cu. ft of air at 300psia and 400°F is cooled to 140°F at constant volume. What is the transferred heat?  a.-120Btu  b. -220Btu  c.-320Btu  d. -420Btu formula: Q= mcv(T2-T1)

Description : Ten cu. ft of air at 300psia and 400°F is cooled to 140°F at constant volume. What is the transferred heat?  a.-120Btu  b. -220Btu  c.-320Btu  d. -420Btu formula: Q= mcv(T2-T1)

Last Answer : -420Btu

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

Description : 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

Last Answer : 28.6 psia

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

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

A certain gas with cp = 0.529Btu/lb°R and R = 96.2ft/lbºR expands from 5 ft and 80ºF to 15 ft while the pressure remains constant at 15.5 psia.  a. T2=1.620ºR, ∫H = 122.83 Btu  b. T2 = 2°R, ∫H = 122.83 Btu  c. T2 = 2.620ºR, ∫H = 122.83 Btu  d. T2 = 1°R, ∫H = 122.83 Btu T2= V2(t2)/V1 and ∫H = mcp (T2-T1)

Description : A certain gas with cp = 0.529Btu/lb°R and R = 96.2ft/lbºR expands from 5 ft and 80ºF to 15 ft while the pressure remains constant at 15.5 psia.  a. T2=1.620ºR, ∫H = 122.83 Btu  b. T2 = 2°R, ∫H = 122.83 Btu  c. ... , ∫H = 122.83 Btu  d. T2 = 1°R, ∫H = 122.83 Btu T2= V2(t2)/V1 and ∫H = mcp (T2-T1)

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

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

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

A certain gas, with cp = 0.529Btu/ lb. °Rand R = 96.2ft.lb/lb. °R, expands from 5 cu ft and 80°F to 15 cu ft while the pressure remains constant at 15.5psia. Compute for T2.  a.1520°R  b. 1620°R  c. 1720°R  d. 1820°R formula: T2= T1V2/V1

Description : A certain gas, with cp = 0.529Btu/ lb. °Rand R = 96.2ft.lb/lb. °R, expands from 5 cu ft and 80°F to 15 cu ft while the pressure remains constant at 15.5psia. Compute for T2.  a.1520°R  b. 1620°R  c. 1720°R  d. 1820°R formula: T2= T1V2/V1

Last Answer : 1620°R

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

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

Which of the following is the mathematical representation of the Charles’s law?  A. V1/V2= P2/P1  B. V1/T1=V2/T2  C. V1/T2=V2/T1  D. V1/V2=√P2/√P1

Description : Which of the following is the mathematical representation of the Charles’s law?  A. V1/V2= P2/P1  B. V1/T1=V2/T2  C. V1/T2=V2/T1  D. V1/V2=√P2/√P1

Last Answer : V1/T1=V2/T2

Which of the following is the Ideal gas law (equation)?  A. V/T = K  B. V= k*(1/P)  C. P1/T1 = P2/T2  D. PV = nRT

Description : Which of the following is the Ideal gas law (equation)?  A. V/T = K  B. V= k*(1/P)  C. P1/T1 = P2/T2  D. PV = nRT

Last Answer : PV = nRT

Assuming compression is according to the Law PV = C, Calculate the initial volume of the gas at a pressure of 2 bars w/c will occupy a volume of 6m³ when it is compressed to a pressure of 42 Bars.  a) 130m³  b) 136m³  c) 120m³  d) 126m³ Formula: P1V1/T1 =P2V2/T2

Description : Assuming compression is according to the Law PV = C, Calculate the initial volume of the gas at a pressure of 2 bars w/c will occupy a volume of 6m³ when it is compressed to a pressure of 42 Bars.  a) 130m³  b) 136m³  c) 120m³  d) 126m³ Formula: P1V1/T1 =P2V2/T2

Last Answer : 126m³

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)

Description : 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)

Last Answer : 339.4 K

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

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

A gas having a volume of100 ft³ at 27ºC is expanded to 120 ft³by heated at constant pressure to what temperature has it been heated to have this new volume?  a. 87°C  b. 85°C  c. 76°C  d. 97°C t2= T2–T1

Description : A gas having a volume of100 ft³ at 27ºC is expanded to 120 ft³by heated at constant pressure to what temperature has it been heated to have this new volume?  a. 87°C  b. 85°C  c. 76°C  d. 97°C t2= T2–T1

Last Answer : 87°C

The heat supplied to the gaS at constant volume is (where m = Mass of gas, cv = Specific heat at constant volume, cp = Specific heat at constant pressure, T2 – T1 = Rise in temperature, and R = Gas constant)  A. mR(T2 – T1)  B. mcv(T2 – T1)  C. mcp(T2 – T1)  D. mcp(T2 + T1)

Description : The heat supplied to the gaS at constant volume is (where m = Mass of gas, cv = Specific heat at constant volume, cp = Specific heat at constant pressure, T2 – T1 = Rise in temperature, and R = Gas constant)  A. mR(T2 – T1)  B. mcv(T2 – T1)  C. mcp(T2 – T1)  D. mcp(T2 + T1)

Last Answer : Answer: B

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

Calculate the entropy of steam at 60psiawith a quality of 0.8  A. 0.4274 BTU/lbm-˚R  B. 0.7303 BTU/lbm-˚R  C. 1.1577 BTU/lbm-˚R  D. 1.2172 BTU/lbm-˚R Formula: fromthe steamtable at 60 psia: sƒ = 0.4274 BTU/lbm-˚R sƒg = 1.2172 BTU/lbm-˚R) s = sƒ + x sƒg where x = is the quality

Description : Calculate the entropy of steam at 60psiawith a quality of 0.8  A. 0.4274 BTU/lbm-˚R  B. 0.7303 BTU/lbm-˚R  C. 1.1577 BTU/lbm-˚R  D. 1.2172 BTU/lbm-˚R Formula: fromthe steamtable at 60 psia: sƒ = 0.4274 BTU/lbm-˚R sƒg = 1.2172 BTU/lbm-˚R) s = sƒ + x sƒg where x = is the quality

Last Answer : 1.1577 BTU/lbm-˚R

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

What is the equation for the work done by a constant temperature system?  A. W = mRTln(V2-V1)  B. W = mR( T2-T1 ) ln( V2/V1)  C. W = mRTln (V2/V1)  D. W = RT ln (V2/V1) Formula : W=∫ pdV lim1,2 ∫ = mRT / V

Description : What is the equation for the work done by a constant temperature system?  A. W = mRTln(V2-V1)  B. W = mR( T2-T1 ) ln( V2/V1)  C. W = mRTln (V2/V1)  D. W = RT ln (V2/V1) Formula : W=∫ pdV lim1,2 ∫ = mRT / V

Last Answer : W = mRTln (V2/V1)

The gas in a constant gas thermometer cooled to absolute zero would have _________.  a. no volume  b. no pressure  c. zero temperature at all scales  d. none of the above

Description : The gas in a constant gas thermometer cooled to absolute zero would have _________.  a. no volume  b. no pressure  c. zero temperature at all scales  d. none of the above

Last Answer : no pressure

Find ∫ for steam at 100 psia and 600°F.If h = 1329.6 and v = 6.216  a. 1214 Btu / lb  b. 1234 Btu /lb  c. 1342 Btu / lb  d. 1324 Btu /lb formula: ∫ = h– pv/ J

Description : Find ∫ for steam at 100 psia and 600°F.If h = 1329.6 and v = 6.216  a. 1214 Btu / lb  b. 1234 Btu /lb  c. 1342 Btu / lb  d. 1324 Btu /lb formula: ∫ = h– pv/ J

Last Answer : 1214Btu / lb

Steam at 1000 lbf/ft^2 pressure and 300˚R has specific volume of 6.5 ft^3/lbm and a specific enthalpy of 9800 lbf-ft/lbm. Find the internal energy per pound mass of steam.  A.2500 lbf-ft/lbm  B.3300 lbf-ft/lbm  C.5400 lbf-ft/lbm  D.6900 lbf-ft/lbm Formula: h= u+ pV u= h– pV

Description : Steam at 1000 lbf/ft^2 pressure and 300˚R has specific volume of 6.5 ft^3/lbm and a specific enthalpy of 9800 lbf-ft/lbm. Find the internal energy per pound mass of steam.  A.2500 lbf-ft/lbm  B.3300 lbf-ft/lbm  C.5400 lbf-ft/lbm  D.6900 lbf-ft/lbm Formula: h= u+ pV u= h– pV

Last Answer : 3300 lbf-ft/lbm

_________ is the temperature to which the air must be cooled, at constant pressure, to produce saturation.  a. relative humidity  b. triple point temperature  c. dew point  d. critical point

Description : _________ is the temperature to which the air must be cooled, at constant pressure, to produce saturation.  a. relative humidity  b. triple point temperature  c. dew point  d. critical point

Last Answer : dew point

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

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

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

Last Answer : All of the above

A car whose mass is 2 metric tons is accelerated uniformly from stand hill to 100 kmph in 5 sec. Find the driving force in Newton’s.  a. 11,120 N  b. 11,320 N  c. 11,420 N  d. 11520 N formula: F= ma / k

Description : A car whose mass is 2 metric tons is accelerated uniformly from stand hill to 100 kmph in 5 sec. Find the driving force in Newton’s.  a. 11,120 N  b. 11,320 N  c. 11,420 N  d. 11520 N formula: F= ma / k

Last Answer : 11,120N

The value for the ΔU of a system is -120 J. If the system is known to have absorbed 420 J of heat, how much work was done?  a. -540 J  b. -640 J  c. -740 J  d. -840 J formula: ΔU = q +w

Description : The value for the ΔU of a system is -120 J. If the system is known to have absorbed 420 J of heat, how much work was done?  a. -540 J  b. -640 J  c. -740 J  d. -840 J formula: ΔU = q +w

Last Answer : -540 J

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

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

The flow energy of 5 ft3 of a fluid passing a boundary to a system is 80,000 ft-lb. Determine the pressure at this point.  a. 222 psi  b. 333 psi  c. 444 psi  d. 111 psi formula: Ef= pV

Description : The flow energy of 5 ft3 of a fluid passing a boundary to a system is 80,000 ft-lb. Determine the pressure at this point.  a. 222 psi  b. 333 psi  c. 444 psi  d. 111 psi formula: Ef= pV

Last Answer : 111 psi

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

Description : 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

Last Answer : 33.9 ft

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

Last Answer : 1

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

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

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)}

An ideal gas of volume 1liter and pressure 10 bar undergoes a quasistatic adiabatic expansion until the pressure drops to 1 bar. Assume γ to be 1.4 what is the final volume?  a. 3.18 l  b. 4.18 l  c. 5.18 l  d. 6.18 l

Description : An ideal gas of volume 1liter and pressure 10 bar undergoes a quasistatic adiabatic expansion until the pressure drops to 1 bar. Assume γ to be 1.4 what is the final volume?  a. 3.18 l  b. 4.18 l  c. 5.18 l  d. 6.18 l

Last Answer : 5.18 l

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

____________is the difference between the saturation temperature for the given pressure and the actual sub cooled liquid temperature.  a) Degrees of Superheat, °SH  b) Degrees of Sub cooled, °SB  c) Both a and b  d) None of the above

Description : ____________is the difference between the saturation temperature for the given pressure and the actual sub cooled liquid temperature.  a) Degrees of Superheat, °SH  b) Degrees of Sub cooled, °SB  c) Both a and b  d) None of the above

Last Answer : Degrees of Sub cooled, °SB

_____________ is the difference between the actual temperature of superheated vapor and the saturation temperature for the existing pressure.  a) Degrees of Superheat, °SH  b) Degrees of Sub cooled, °SB  c) Both a and b  d) None of the above

Description : _____________ is the difference between the actual temperature of superheated vapor and the saturation temperature for the existing pressure.  a) Degrees of Superheat, °SH  b) Degrees of Sub cooled, °SB  c) Both a and b  d) None of the above

Last Answer : Degrees of Superheat, °SH

__________________ is a liquid at the saturations which has temperature equal to the boiling point corresponding to the existing pressure.  a. Saturation Temperature  b. Sub cooled Liquid  c. Compressed Liquid  d. Saturated Liquid

Description : __________________ is a liquid at the saturations which has temperature equal to the boiling point corresponding to the existing pressure.  a. Saturation Temperature  b. Sub cooled Liquid  c. Compressed Liquid  d. Saturated Liquid

Last Answer : Saturated Liquid

_________________ is one which has a pressure higher than the saturation pressure corresponding to the existing temperature.  a. Saturation Temperature  b. Sub cooled Liquid  c. Compressed Liquid  d. Saturated Liquid

Description : _________________ is one which has a pressure higher than the saturation pressure corresponding to the existing temperature.  a. Saturation Temperature  b. Sub cooled Liquid  c. Compressed Liquid  d. Saturated Liquid

Last Answer : Compressed Liquid

________________ is one which has a temperature lower than the saturation temperature corresponding to the existing pressure.  a. Saturation Temperature  b. Sub cooled Liquid  c. Compressed Liquid  d. Saturated Liquid

Description : ________________ is one which has a temperature lower than the saturation temperature corresponding to the existing pressure.  a. Saturation Temperature  b. Sub cooled Liquid  c. Compressed Liquid  d. Saturated Liquid

Last Answer : Sub cooled Liquid

In thermodynamics, a throttling process, also called a _________, is a type of isenthalpic process where a liquid or gas is cooled as it passes from a higher pressure state to a lower pressure state.  a. Rankine Process  b. Carnot Cycle  c. Joule-Thomson process  d. Refrigeration process

Description : In thermodynamics, a throttling process, also called a _________, is a type of isenthalpic process where a liquid or gas is cooled as it passes from a higher pressure state to a lower pressure state.  a. Rankine Process  b. Carnot Cycle  c. Joule-Thomson process  d. Refrigeration process

Last Answer : Joule-Thomson process

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

If the value of n = 0 in the equation pvn = C, then the process is called  A. constant volume process  B. adiabatic process  C. constant pressure process  D. isothermal process

Description : If the value of n = 0 in the equation pvn = C, then the process is called  A. constant volume process  B. adiabatic process  C. constant pressure process  D. isothermal process

Last Answer : Answer: C

The volume of a gas under standard atmospheric pressure & 76 cmHg is 200m³. What is the volume when pressure is 80 cmHg if the temperature is unchanged?  a) 180 in³  b) 170 in³  c) 160 in³  d) 190 in³ Formula: P2V2 = P1V1

Description : The volume of a gas under standard atmospheric pressure & 76 cmHg is 200m³. What is the volume when pressure is 80 cmHg if the temperature is unchanged?  a) 180 in³  b) 170 in³  c) 160 in³  d) 190 in³ Formula: P2V2 = P1V1

Last Answer : 190 in³

For heat engine operating between two temperatures (T1>T2), what is the maximum efficiency attainable?  A. Eff = 1 – (T2/T1)  B. Eff = 1 - (T1/T2)  C. Eff = T1 - T2  D. Eff = 1 - (T2/T1)^2

Description : For heat engine operating between two temperatures (T1>T2), what is the maximum efficiency attainable?  A. Eff = 1 – (T2/T1)  B. Eff = 1 - (T1/T2)  C. Eff = T1 - T2  D. Eff = 1 - (T2/T1)^2

Last Answer : Eff = 1 – (T2/T1)