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

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
by

1 Answer

Answer :

14026.5 m³
by

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

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According to Avogadro’s law  A. the product of the gas constant and the molecular mass of an ideal gas is constant  B. the sum of partial pressure of the mixture of two gases is sum of the two  C. equal volumes of all gases, at the same temperature and pressure, contain equal number of molecules  D. all of the above

Description : According to Avogadro's law  A. the product of the gas constant and the molecular mass of an ideal gas is constant  B. the sum of partial pressure of the mixture of two gases is sum of the ... all gases, at the same temperature and pressure, contain equal number of molecules  D. all of the above

Last Answer : Answer: C

Gas turbine cycle consists of  (a) two isothermals and two isentropics  (b) two isentropics and two constant volumes  (c) two isentropics, one constant volume and one constant pressure  (d) two isentropics and two constant pressures  (e) none of the above.

Description : Gas turbine cycle consists of  (a) two isothermals and two isentropics  (b) two isentropics and two constant volumes  (c) two isentropics, one constant volume and one constant pressure  (d) two isentropics and two constant pressures  (e) none of the above.

Last Answer : Answer : d

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

The flow energy of 150 L of a fluid passing a boundary to a system is 110 kJ. Determine the pressure at this point  a. 733.33 kPa  b. 833.33 kPa  c. 933.33 kPa  d. 633.33 kPa

Description : The flow energy of 150 L of a fluid passing a boundary to a system is 110 kJ. Determine the pressure at this point  a. 733.33 kPa  b. 833.33 kPa  c. 933.33 kPa  d. 633.33 kPa

Last Answer : 733.33 kPa

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

Brayton cycle consists’ of following four processes  (a) two isothermals and two isentropics  (b) two isentropics and two constant volumes  (c) two isentropics, one constant volume and one constant pressure  (d) two isentropics and two constant pres-sures  (e) none of the above.

Description : Brayton cycle consists’ of following four processes  (a) two isothermals and two isentropics  (b) two isentropics and two constant volumes  (c) two isentropics, one constant volume and one constant pressure  (d) two isentropics and two constant pres-sures  (e) none of the above.

Last Answer : Answer : d

Diesel cycle consists of following four processes  (a) two isothermals and two isentropics  (b) two isentropics, and two constant volumes.  (c) two isentropics, one constant volume and one constant pressure  (d) two isentropics and two constant pressures  (e) none of the above.

Description : Diesel cycle consists of following four processes  (a) two isothermals and two isentropics  (b) two isentropics, and two constant volumes.  (c) two isentropics, one constant volume and one constant pressure  (d) two isentropics and two constant pressures  (e) none of the above.

Last Answer : Answer : c

Otto cycle consists of following four processes  (a) two isothermals and two isentropics  (b) two isentropics and two constant volumes  (c) two isentropics, one constant volume and one constant pressure  (d) two isentropics and two constant pres-sures  (e) none of the above.

Description : Otto cycle consists of following four processes  (a) two isothermals and two isentropics  (b) two isentropics and two constant volumes  (c) two isentropics, one constant volume and one constant pressure  (d) two isentropics and two constant pres-sures  (e) none of the above.

Last Answer : Answer : b

The volume of a gas is directly proportional to the number of molecules of the gas.  a. Ideal gas law  b. Boyle-Mariotte Law  c. Avogadro’s Hypothesis  d. Gay-Lussac’s Law of combining Volumes

Description : The volume of a gas is directly proportional to the number of molecules of the gas.  a. Ideal gas law  b. Boyle-Mariotte Law  c. Avogadro’s Hypothesis  d. Gay-Lussac’s Law of combining Volumes

Last Answer : Avogadro’s Hypothesis

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

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

An engine operates between temperatures of 900°Kandr2 and another engine between T2 and 400°K For both to do equal work, value of T2 will be  (a) 650°K  (b) 600°K  (c) 625°K  (d) 700°K  (e) 750°K.

Description : An engine operates between temperatures of 900°Kandr2 and another engine between T2 and 400°K For both to do equal work, value of T2 will be  (a) 650°K  (b) 600°K  (c) 625°K  (d) 700°K  (e) 750°K.

Last Answer : Answer : a