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

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
by

1 Answer

Answer :

28.6 psia
by

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Description : The expression, ∆G = nRT. ln(P2/P1), gives the free energy change (A) With pressure changes at constant temperature (B) Under reversible isothermal volume change (C) During heating of an ideal gas (D) During cooling of an ideal gas

Last Answer : (A) With pressure changes at constant temperature

An automobile tire has a gauge pressure of 200 kpa at 0°C assuming no air leaks and no change of volume of the tire, what is the gauge pressure at 35ºC.  a. 298.645  b. 398.109  c. 291.167  d. 281.333 Pg = Pabs - Patm

Description : An automobile tire has a gauge pressure of 200 kpa at 0°C assuming no air leaks and no change of volume of the tire, what is the gauge pressure at 35ºC.  a. 298.645  b. 398.109  c. 291.167  d. 281.333 Pg = Pabs - Patm

Last Answer : 298.645

Two lines L1 and L2 having Parametric equations are P1=[3 4 7]+u[2 2 -6] and P2=[15 -2]+u[1 4 2]. Tangent vector for line L1 a.2i+2j-6k b.2i+2j+6k c.2i-2j-6k d.6-2j-2k

Description : Two lines L1 and L2 having Parametric equations are P1=[3 4 7]+u[2 2 -6] and P2=[15 -2]+u[1 4 2]. Tangent vector for line L1 a.2i+2j-6k b.2i+2j+6k c.2i-2j-6k d.6-2j-2k

Last Answer : a.2i+2j-6k

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

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

Description : 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

Last Answer : 217.17 kpa

Water (specific heat cv= 4.2 kJ/ kg ∙ K ) is being heated by a 1500 W h eater. What is the rate of change in temperature of 1kg of the water?  A. 0.043 K/s  B. 0.179 K/s  C. 0.357 K/s  D. 1.50 K/s Formula: Q = mcv ( T)

Description : Water (specific heat cv= 4.2 kJ/ kg ∙ K ) is being heated by a 1500 W h eater. What is the rate of change in temperature of 1kg of the water?  A. 0.043 K/s  B. 0.179 K/s  C. 0.357 K/s  D. 1.50 K/s Formula: Q = mcv ( T)

Last Answer : 0.179 K/s

In a 2-D CAD package, clockwise circular arc of radius, 5, specified from P1 (15,10)to P2 (10,15)will have its center at a.(10, 10) b.(15, 10) c.(15, 15) d.(10, 15)

Description : In a 2-D CAD package, clockwise circular arc of radius, 5, specified from P1 (15,10)to P2 (10,15)will have its center at a.(10, 10) b.(15, 10) c.(15, 15) d.(10, 15)

Last Answer : a.(10, 10)

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 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

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