The value of gas constant (R) in S. I. units is  A. 0.287 J/kgK  B. 2.87 J/kgK  C. 28.7 J/kgK  D. 287 J/kgK

The value of gas constant (R) in S. I. units is  A. 0.287 J/kgK  B. 2.87 J/kgK  C. 28.7 J/kgK  D. 287 J/kgK
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Answer: D
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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

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

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

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

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

Steam is throttled to 0.1 MPa with 20 degrees of superheat. (a) What is the quality of throttled steam if its pressure is 0.75 MPa (b) What is the enthalpy of the process?  a) 97.6%,2713 kJ/kg  b) -97.6%, 2713 kJ/kg  c) 87.6%,3713 kJ/kg  d) -87.6%, 3713 kJ/kg

Description : Steam is throttled to 0.1 MPa with 20 degrees of superheat. (a) What is the quality of throttled steam if its pressure is 0.75 MPa (b) What is the enthalpy of the process?  a) 97.6%,2713 kJ/kg  b) -97.6%, 2713 kJ/kg  c) 87.6%,3713 kJ/kg  d) -87.6%, 3713 kJ/kg

Last Answer : 97.6%,2713 kJ/kg

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

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

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

Determine the amount of heat required to produce 1Kg of steam at a pressure of 7 bar, at a temperature 29 degree Celsius, under the following . (i) When the steam is wet having dryness fraction 0.87 (ii) When the steam is dry saturated Assume specific heat = 2.35KJ/KgK

Description : Determine the amount of heat required to produce 1Kg of steam at a pressure of 7 bar, at a temperature 29 degree Celsius, under the following . (i) When the steam is wet having dryness fraction 0.87 (ii) When the steam is dry saturated Assume specific heat = 2.35KJ/KgK 

Last Answer : P 7bar T 29ᵒC x=0.9 m =1kg C p =2.35 KJ/KgK At pressure 7 bar saturation Temp T sat =165 ᵒC h f =697.2 KJ/Kg h fg= 2064.8 KJ/Kg i) Heat required when steam is wet ... ) When the stem is dry saturated hg=hf+hfg =2762 kj Heat actually required = 2762-121.8 = 2640.2 kj

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

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

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 equation Pv = RT, the constant of proportionality R is known as ______.  A. Universal gas constant  B. Gas constant  C. Ideal gas factor  D. Gas index

Description : In the equation Pv = RT, the constant of proportionality R is known as ______.  A. Universal gas constant  B. Gas constant  C. Ideal gas factor  D. Gas index

Last Answer : Gas constant

The gas constant (R) is equal to the __________ of two specific heats.  A. sum  B. difference  C. product  D. ratio

Description : The gas constant (R) is equal to the __________ of two specific heats.  A. sum  B. difference  C. product  D. ratio

Last Answer : Answer: B

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

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

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

Description : 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 ... °C  b. 8.9°C  c. 9.9°C  d. 10.9°C formula: ΔQ = m c ΔT

Last Answer : 7.9 °C

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 thick slices of bread, when completely oxidized by the body, can supply 200,000 cal of heat. How much work is this equivalent to?  a) 4,190,000 joules  b) 8,390,000 joules  c) 839,000 joules  d) 419 000 joules Formula: J =Work/Heat J = mechanical equivalent of heat whose value is 4.19 joules/calorie

Description : Two thick slices of bread, when completely oxidized by the body, can supply 200,000 cal of heat. How much work is this equivalent to?  a) 4,190,000 joules  b) 8,390,000 joules  c) 839, ... d) 419 000 joules Formula: J =Work/Heat J = mechanical equivalent of heat whose value is 4.19 joules/calorie

Last Answer : 419 000 joules

How many joules of work is the equivalent of 15000 cal of heat?  a) 62850 joules  b) 3579.95 joules  c) 14995.81 joules  d) 15004.19 joules Formula: J =Work/Heat J = mechanical equivalent of heat whose value is 4.19 joules/calorie

Description : How many joules of work is the equivalent of 15000 cal of heat?  a) 62850 joules  b) 3579.95 joules  c) 14995.81 joules  d) 15004.19 joules Formula: J =Work/Heat J = mechanical equivalent of heat whose value is 4.19 joules/calorie

Last Answer : 62850 joules

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

The compressibility factor, x, is used for predicting the behavior of nonideal gases. How is the compressibility ty factor defined relative to an ideal gas? (subscript “c”refers to critical value)  A. z = P / Pc  B. z = pV/ RT  C. z = T /Tc  D. z = RT / pV Hint: for an real gases the compressibility factor, x, is an dimensionless constant given by pV=zRT. Therefore z = pV / RT

Description : The compressibility factor, x, is used for predicting the behavior of nonideal gases. How is the compressibility ty factor defined relative to an ideal gas? (subscript c refers to critical value)  A. ... compressibility factor, x, is an dimensionless constant given by pV=zRT. Therefore z = pV / RT

Last Answer : z = pV/ RT

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

The unit of power in S.I. units is  (a) newton  (b) pascal  (c) erg  (d) watt  (e) joule.

Description : The unit of power in S.I. units is  (a) newton  (b) pascal  (c) erg  (d) watt  (e) joule.

Last Answer : Answer : d

The unit of pressure in S.I. units is  (a) kg/cm2  (b) mm of water column  (c) pascal  (d) dynes per square cm  (e) bars

Description : The unit of pressure in S.I. units is  (a) kg/cm2  (b) mm of water column  (c) pascal  (d) dynes per square cm  (e) bars

Last Answer : Answer : c

The unit of energy in S.I. units is  (a) watt  (b) joule  (c) joule/s  (d) joule/m  (e) joule m.

Description : The unit of energy in S.I. units is  (a) watt  (b) joule  (c) joule/s  (d) joule/m  (e) joule m.

Last Answer : Answer : b

The unit of time in S.I. units is  (a) second  (b) minute  (c) hour  (d) day  (e) year

Description : The unit of time in S.I. units is  (a) second  (b) minute  (c) hour  (d) day  (e) year

Last Answer : Answer : a

The unit of length in S.I. units is  (a) meter  (b) centimeter  (c) kilometer  (d) millimeter.

Description : The unit of length in S.I. units is  (a) meter  (b) centimeter  (c) kilometer  (d) millimeter.

Last Answer : Answer : a

The unit of mass in S.I. units is  (a) kilogram  (b) gram  (c) tonne  (d) quintal  (e) newton

Description : The unit of mass in S.I. units is  (a) kilogram  (b) gram  (c) tonne  (d) quintal  (e) newton

Last Answer : Answer : a

The unit of temperature in S.I. units is  (a) Centigrade  (b) Celsius  (c) Fahrenheit  (d) Kelvin  (e) Rankine.

Description : The unit of temperature in S.I. units is  (a) Centigrade  (b) Celsius  (c) Fahrenheit  (d) Kelvin  (e) Rankine.

Last Answer : Answer : d

In the above problem, find the potential energy of the mass with respect to datum. (Formula: P = mgz/k )  a. 4875 j  b. 0.51 j  c. 0.46 j  d. None of the above

Description : In the above problem, find the potential energy of the mass with respect to datum. (Formula: P = mgz/k )  a. 4875 j  b. 0.51 j  c. 0.46 j  d. None of the above

Last Answer : 4875 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 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

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

A simple steam engine receives steam from the boiler at 180˚C and exhausts directly into the air at 100˚C. What is the upper limit of its efficiency?  a. 11.28 %  b. 36.77 %  c. 20.36 %  d. 17.66 %

Description : A simple steam engine receives steam from the boiler at 180˚C and exhausts directly into the air at 100˚C. What is the upper limit of its efficiency?  a. 11.28 %  b. 36.77 %  c. 20.36 %  d. 17.66 %

Last Answer : 17.66 %

The SI unit of pressure  a. Pa  b. N  c. J  d. None of the above

Description : The SI unit of pressure  a. Pa  b. N  c. J  d. None of the above

Last Answer : Pa

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

The concrete roof of a house is 10 m by 8 m and 10 cm thick (4"). Estimate the total heat the roof would absorb over the 12 day?  a. 1.3 x 108 J  b 2.3 x 108 J  c. 3.3 x 108 J  d. 4.3 x 108 J formula: ΔQ = ΔQ/Δtx Δt

Description : The concrete roof of a house is 10 m by 8 m and 10 cm thick (4"). Estimate the total heat the roof would absorb over the 12 day?  a. 1.3 x 108 J  b 2.3 x 108 J  c. 3.3 x 108 J  d. 4.3 x 108 J formula: ΔQ = ΔQ/Δtx Δt

Last Answer : 1.3 x 108 J

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

If a system absorbs 500 cal of heat at the same time does 400J of work, find the change in internal energy of the system.  a. 1400 J  b. 1700 J  c. 1900 J  d. 1500 J

Description : If a system absorbs 500 cal of heat at the same time does 400J of work, find the change in internal energy of the system.  a. 1400 J  b. 1700 J  c. 1900 J  d. 1500 J

Last Answer : 1700 J

What is the amount of radiant energy received each second over each square meter that is at right angles to the sun’s rays at the top of the atmosphere?  a. 1400 J  b. 6000 J  c. 10000 J  d. 800 J

Description : What is the amount of radiant energy received each second over each square meter that is at right angles to the sun’s rays at the top of the atmosphere?  a. 1400 J  b. 6000 J  c. 10000 J  d. 800 J

Last Answer : 1400 J

What is the SI unit of specific heat capacity?  A. J/kg  B. J/kg °F ∙  C. J/kg °C ∙  D. J/°C

Description : What is the SI unit of specific heat capacity?  A. J/kg  B. J/kg °F ∙  C. J/kg °C ∙  D. J/°C

Last Answer : J/kg °C

What is the specific heat capacity of water in J/kg °C? ∙  A. 4581  B. 4185  C. 4518  D. 4815

Description : What is the specific heat capacity of water in J/kg °C? ∙  A. 4581  B. 4185  C. 4518  D. 4815

Last Answer : 4185

Kinetic energy of the molecules in terms of absolute temperature (T) is proportional to  (a) T  (b) j  (c) J2  (d) Vr  (e) 1/Vr.

Description : Kinetic energy of the molecules in terms of absolute temperature (T) is proportional to  (a) T  (b) j  (c) J2  (d) Vr  (e) 1/Vr.

Last Answer : Answer : a

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 volume of a gas under constant pressure increases or decrease with temperature.  a. Gay- Lussac’s Law  b. Ideal Gas Law  c. Charles’ Law  d. Boyle’s Law

Description : The volume of a gas under constant pressure increases or decrease with temperature.  a. Gay- Lussac’s Law  b. Ideal Gas Law  c. Charles’ Law  d. Boyle’s Law

Last Answer : Charles’ Law