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

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
by

1 Answer

Answer :

1620 °R
by

Related questions

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

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

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

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

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

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

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

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

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 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

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)

Calculate: a. Mass flow rate in lb/hr. b. The velocity at section 2 in fps  a. 800,000lb/hr;625ft/s  b. 900,000lb/hr;625 ft/s  c. 888,000lb/hr;269 ft/s  d. 700,000lb/hr;269 ft/s m = A1V!/V1

Description : Calculate: a. Mass flow rate in lb/hr. b. The velocity at section 2 in fps  a. 800,000lb/hr;625ft/s  b. 900,000lb/hr;625 ft/s  c. 888,000lb/hr;269 ft/s  d. 700,000lb/hr;269 ft/s m = A1V!/V1

Last Answer : 900,000 lb/hr;625 ft/s

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

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

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

In the above problem, compute for the mass. (Formula: m = p1V1 / RT1)  a. 0.2148 lb  b. 0.2134 lb  c. 0.1248 lb  d. None of the above

Description : In the above problem, compute for the mass. (Formula: m = p1V1 / RT1)  a. 0.2148 lb  b. 0.2134 lb  c. 0.1248 lb  d. None of the above

Last Answer : 0.2148 lb

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

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

A fluid flows in a steady manner between two section in a flow line at section 1: A 1 = 1ft², V1 = 100fpm, volume1 of 4ft³/lb. at sec2: A2 = 2 ft², p= 0.20 lb/ft³ calculate the velocity at section 2.  a. 625 fpm  b. 567 fpm  c. 356 fpm  d. None of the above

Description : A fluid flows in a steady manner between two section in a flow line at section 1: A 1 = 1ft², V1 = 100fpm, volume1 of 4ft³/lb. at sec2: A2 = 2 ft², p= 0.20 lb/ft³ calculate the velocity at section 2.  a. 625 fpm  b. 567 fpm  c. 356 fpm  d. None of the above

Last Answer : 625 fpm

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

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

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

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

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

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

A System has a temperature of 250°F. Convert this Value to °R?  a. 740°R  b.730°R  c. 720°R  d. 710°R formula: °R= °F + 460

Description : A System has a temperature of 250°F. Convert this Value to °R?  a. 740°R  b.730°R  c. 720°R  d. 710°R formula: °R= °F + 460

Last Answer : 710°R

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

What mass of nitrogen is contained in a10 ft3 vessel at a pressure of 840atm and 820°R? Make a computation by using ideal gas equation.  a. 194lb  b. 214lb  c. 394 lb  d. 413lb formula: m=pV /RT

Description : What mass of nitrogen is contained in a10 ft3 vessel at a pressure of 840atm and 820°R? Make a computation by using ideal gas equation.  a. 194lb  b. 214lb  c. 394 lb  d. 413lb formula: m=pV /RT

Last Answer : 394 lb

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 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

The value of specific heat at constant pressure (cp) is __________ that of at constant volume (cv).  A. less than  B. equal to  C. more than

Description : The value of specific heat at constant pressure (cp) is __________ that of at constant volume (cv).  A. less than  B. equal to  C. more than

Last Answer : Answer: C

The ratio of specific heat at constant pressure (Cp) and specific heat at constant volume (cv) is  A. equal to one  B. less than one  C. greater than one  D. none of these

Description : The ratio of specific heat at constant pressure (Cp) and specific heat at constant volume (cv) is  A. equal to one  B. less than one  C. greater than one  D. none of these

Last Answer : Answer: C

The gas constant is equal to  a. Cp – Cv  b. Cp + Cv  c. Cp – Cv + k  d. None of the above

Description : The gas constant is equal to  a. Cp – Cv  b. Cp + Cv  c. Cp – Cv + k  d. None of the above

Last Answer : Cp – Cv

Characteristic gas constant of a gas is equal to  (a) C/Cv  (b) Cv/Cp  (c) Cp – Cv  (d) Cp + Cv  (e) Cp x Cv

Description : Characteristic gas constant of a gas is equal to  (a) C/Cv  (b) Cv/Cp  (c) Cp – Cv  (d) Cp + Cv  (e) Cp x Cv

Last Answer : Answer : c

For a certain gas R = 320 J/kg.K and cv= 0.84kJ/kg.K. Find k?  a. 1.36  b. 1.37  c. 1.38  d. 1.39 formula: k= R / cv+1

Description : For a certain gas R = 320 J/kg.K and cv= 0.84kJ/kg.K. Find k?  a. 1.36  b. 1.37  c. 1.38  d. 1.39 formula: k= R / cv+1

Last Answer : 1.38

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

The value of gas constant 'R' is (A) 1.987 cal/gm mole °K (B) 1.987 BTU/lb. mole °R (C) Both (A) and (B)(D) Neither (A) nor (B)

Description : The value of gas constant 'R' is (A) 1.987 cal/gm mole °K (B) 1.987 BTU/lb. mole °R (C) Both (A) and (B) (D) Neither (A) nor (B)

Last Answer : (C) Both (A) and (B)

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

876 °R = _____ °F  a. 335  b. 416  c. 400  d. None of the above

Description : 876 °R = _____ °F  a. 335  b. 416  c. 400  d. None of the above

Last Answer : 416

746 °R = ______ °F  a. 254  b. 345  c. 286  d. None of the above

Description : 746 °R = ______ °F  a. 254  b. 345  c. 286  d. None of the above

Last Answer : 286

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

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

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.

Description : 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 ... , if temperature is kept constant  (e) remains constant,if volume and temperature are kept constant.

Last Answer : Answer : c

General gas equation is  (a) PV=nRT  (b) PV=mRT  (d) PV = C  (c) PV=KiRT  (e) Cp-Cv = Wj

Description : General gas equation is  (a) PV=nRT  (b) PV=mRT  (d) PV = C  (c) PV=KiRT  (e) Cp-Cv = Wj

Last Answer : Answer : b

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