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# Multiple choice question for engineering

## Set 1

1. A reactor is supplied with 5 moles each of CH3OH and O2, and the following reactions occur, CH3OH + 1/2O2 -> CH2O + H2O and then CH2O + 1/2O2 -> CO + H2O, the conversion of methanol is 80% in first reaction, and conversion of CH2O is 50% in second reaction. How many moles of CO are produced?
a) 1
b) 2
c) 3
d) 4

Answer: b [Reason:] Extent of reaction E1 = (-0.8)/(-1) = 0.8, E2 = (-0.5)/(-1) = 0.5, => moles of CO = (0.8*5)*0.5 = 2.

2. A reactor is supplied with 5 moles each of CH3OH and O2, and the following reactions occur, CH3OH + 1/2O2 -> CH2O + H2O and then CH2O + 1/2O2 -> CO + H2O, the conversion of methanol is 80% in first reaction, and conversion of CH2O is 50% in second reaction. How many moles of H2O are produced?
a) 2
b) 3
c) 5
d) 6

Answer: d [Reason:] Extent of reaction E1 = (-0.8)/(-1) = 0.8, E2 = (-0.5)/(-1) = 0.5, => moles of H2O = (0.8*5) + (0.8*5)*0.5 = 6.

3. A reactor is supplied with 5 moles each of CH3OH and O2, and the following reactions occur, CH3OH + 1/2O2 -> CH2O + H2O and then CH2O + 1/2O2 -> CO + H2O, the conversion of methanol is 80% in first reaction, and conversion of CH2O is 50% in second reaction. How many moles of CH2O came out of the reactor?
a) 1
b) 2
c) 3
d) 4

Answer: b [Reason:] Extent of reaction E1 = (-0.8)/(-1) = 0.8, E2 = (-0.5)/(-1) = 0.5, => moles of CH2O unreacted = 0.8*5 – (0.8*5)*0.5 = 2.

4. A reactor is supplied with 5 moles each of CH3OH and O2, and the following reactions occur, CH3OH + 1/2O2 -> CH2O + H2O and then CH2O + 1/2O2 -> CO + H2O, the conversion of methanol is 80% in first reaction, and conversion of CH2O is 50% in second reaction. How many moles of O2 are there in the products?
a) 1.75
b) 2
c) 3.25
d) 5.5

Answer: a [Reason:] Extent of reaction E1 = (-0.8)/(-1) = 0.8, E2 = (-0.5)/(-1) = 0.5, => moles of O2 = (2.5 – 2.5*0.8) + (2.5 – 2.5*0.5) = 1.75.

5. A reactor is supplied with 5 moles each of CH3OH and O2, and the following reactions occur, CH3OH + 1/2O2 -> CH2O + H2O and then CH2O + 1/2O2 -> CO + H2O, the conversion of methanol is 80% in first reaction, and conversion of CH2O is 50% in second reaction. How many moles of CH3OH are there in the products?
a) 1
b) 2
c) 3
d) 4

Answer: a [Reason:] Extent of reaction E1 = (-0.8)/(-1) = 0.8, E2 = (-0.5)/(-1) = 0.5, => moles of CH3OH = 5 – 5*0.8 = 1.

6. A reactor is supplied with C6H12O6 and the following reactions occur C6H12O6 -> 2C2H5OH + 2CO2 and C6H12O6 –> 2C2H3CO2H + 2H2O, 360 grams of pure glucose is supplied and 44 grams of CO2 was produced, what 90 grams of glucose left unreacted, how many grams of propenoic acid was produced?
a) 18
b) 36
c) 72
d) 144

Answer: d [Reason:] Weight of glucose reacted in first reaction = (44/44)*(1/2)*180 = 90 grams, => (360 – 90 – 90) = 180 grams of glucose reacted in second reaction, => weight of propenoic acid produced = 2*(180/180)*72 = 144 grams.

7. A reactor is supplied with C6H12O6 and the following reactions occur C6H12O6 -> 2C2H5OH + 2CO2 and C6H12O6 –> 2C2H3CO2H + 2H2O, 360 grams of pure glucose is supplied and 44 grams of CO2 was produced, what 90 grams of glucose left unreacted, how many grams of ethanol was produced?
a) 23
b) 46
c) 92
d) 186

Answer: b [Reason:] Moles of glucose reacted in first reaction = 90/180 = 0.5, => weight of ethanol produced = 2*0.5*46 = 46 grams.

8. A reactor is supplied with C6H12O6 and the following reactions occur C6H12O6 -> 2C2H5OH + 2CO2 and C6H12O6 –> 2C2H3CO2H + 2H2O, 360 grams of pure glucose is supplied and 44 grams of CO2 was produced, what 90 grams of glucose left unreacted, how many grams of water was produced?
a) 9
b) 18
c) 36
d) 54

Answer: a [Reason:] Weight of glucose reacted in first reaction = ½*(66/44)*180 = 135 grams => Weight of glucose reacted in second reaction = 360 – 180 – 135 = 45 grams, => moles of water produced = 2*(45/180)*18 = 9 grams.

9. A reactor is supplied with 5 moles of C4H10, and it undergoes following reactions C4H10 + 4H2O -> 4CO + 9H2 and C4H10 + 6.5O2 -> 4CO2 + 5H2O, if 8 moles of CO2 was produced and 2 moles of C4H10 remains unreacted, what are the number of moles of water produced?
a) 2
b) 5
c) 7.5
d) 10

Answer: d [Reason:] Extent of second reaction = (8 – 0)/4 = 2, => moles of water produced = 5*2 = 10.

10. A reactor is supplied with 5 moles of C4H10, and it undergoes following reactions C4H10 + 4H2O -> 4CO + 9H2 and C4H10 + 6.5O2 -> 4CO2 + 5H2O, if 8 moles of CO2 was produced and 2 moles of C4H10 remains unreacted, what are the number of moles of CO produced?
a) 1
b) 2
c) 3
d) 4

Answer: d [Reason:] Extent of second reaction = (8 – 0)/4 = 2, => moles of C4H10 reacted in second reaction = 2*1 = 2, => moles of C4H10 reacted in first reaction = 5 – 2 – 2 = 1, => extent of first reaction = (0 – 1)/(-1) = 1, => moles of CO produced = 4*1 = 4.

11. A reactor is supplied with 12 moles of C, and it undergoes following reactions C + 0.5O2 -> CO and C + O2 -> CO2, if 4 moles of CO2 was produced and 4 moles of C remains unreacted, what are the number of moles of CO produced?
a) 1
b) 2
c) 3
d) 4

Answer: d [Reason:] Extent of second reaction = (4 – 0)/1 = 4, => moles of C reacted in second reaction = 4*1 = 4, => moles of C reacted in first reaction = 12 – 4 – 4 = 4, => extent of first reaction = (0 – 4)/(-1) = 4, => moles of CO produced = 4*1 = 4.

12. A reactor is supplied with 15 moles of C, and it undergoes following reactions C + 0.5O2 -> CO and C + O2 -> CO2, if 8 moles of CO2 was produced and 3 moles of C remains unreacted, what are the number of moles of CO produced?
a) 1
b) 2
c) 3
d) 4

Answer: d [Reason:] Extent of second reaction = (8 – 0)/1 = 8, => moles of C reacted in second reaction = 8*1 = 8, => moles of C reacted in first reaction = 15 – 8 – 3 = 4, => extent of first reaction = (0 – 4)/(-1) = 4, => moles of CO produced = 4*1 = 4.

13. A reactor is supplied with 25 moles of C, and it undergoes following reactions C + 0.5O2 -> CO and C + O2 -> CO2, if 13 moles of CO2 was produced and 5 moles of C remains unreacted, what are the number of moles of CO produced?
a) 1
b) 3
c) 5
d) 7

Answer: d [Reason:] Extent of second reaction = (13 – 0)/1 = 13, => moles of C reacted in second reaction = 13*1 = 13, => moles of C reacted in first reaction = 25 – 13 – 5 = 7, => extent of first reaction = (0 – 7)/(-1) = 7, => moles of CO produced = 7*1 = 7.

14. A reactor is supplied with 20 moles of C, and it undergoes following reactions C + 0.5O2 -> CO and C + O2 -> CO2, if 10 moles of CO2 was produced and 5 moles of C remains unreacted, what are the number of moles of CO produced?
a) 1
b) 3
c) 5
d) 7

Answer: c [Reason:] Extent of second reaction = (10 – 0)/1 = 10, => moles of C reacted in second reaction = 10*1 = 10, => moles of C reacted in first reaction = 20 – 10 – 5 = 5, => extent of first reaction = (0 – 5)/(-1) = 5, => moles of CO produced = 5*1 = 5.

15. A reactor is supplied with 45 moles of C, and it undergoes following reactions C + 0.5O2 -> CO and C + O2 -> CO2, if 25 moles of CO2 was produced and 6 moles of C remains unreacted, what are the number of moles of CO produced?
a) 6
b) 9
c) 14
d) 17

Answer: c [Reason:] Extent of second reaction = (25 – 0)/1 = 25, => moles of C reacted in second reaction = 25*1 = 25, => moles of C reacted in first reaction = 45 – 25 – 6 = 14, => extent of first reaction = (0 – 14)/(-1) = 14, => moles of CO produced = 14*1 = 14.

## Set 2

Answer question 1 – 5 for the following diagram, the reaction is A -> 2B. 1. What is the value of W?
a) 10 mole/hr
b) 40 mole/hr
c) 50 mole/hr
d) 90 mole/hr

Answer: c [Reason:] W = 10 + 40 = 50 mole/hr.

2. What is the value of P?
a) 5 mole/hr
b) 10 mole/hr
c) 25 mole/hr
d) 40 mole/hr

Answer: b [Reason:] P = 20% of W, => P = 50*20/100 = 10 mole/hr.

3. What is the overall extent of reaction?
a) 5
b) 10
c) 20
d) 25

Answer: a [Reason:] Extent of reaction = (P – 0)/2 = (10 – 0)/2.

4. What is the overall fraction conversion?
a) 10%
b) 20%
c) 80%
d) 100%

Answer: a [Reason:] Overall fraction conversion = (10 – 0)/10*100 = 100%.

5. What is the single pass fraction conversion?
a) 10%
b) 20%
c) 25%
d) 50%

Answer: b [Reason:] Single pass fraction conversion = (50 – 40)/50*100 = 20%.

Answer the question 6 – 10 for the following diagram. The reaction is C2H4 + H2 -> C2H6, the overall conversion rate of C2H4 is 90%, and the single pass conversion rate is 20%. 6. What is the extent of reaction?
a) 3.6
b) 4.8
c) 6.4
d) 7.2

Answer: a [Reason:] Extent of reaction = 4*0.9/1 = 3.6.

7. What are the moles of ethane in products?
a) 2.4
b) 3.6
c) 6.4
d) 7.2

Answer: b [Reason:] Moles of C2H6 = 3.6*1 = 3.6.

8. What is the value of P?
a) 3.6 mole/hr
b) 5.4 mole/hr
c) 6.4 mole/hr
d) 7.8 mole/hr

Answer: c [Reason:] The extent of reaction = (4*0.9)/1 = 3.6, => moles of C2H6 = 3.6, moles of H2 = 6 – 3.6 = 2.4, moles of C2H4 = 4 – 3.6 = 0.4, => P = 3.6 + 2.4 + 0.4 = 6.4 moles/hr.

9. What is the value of R?
a) 18.4 mole/hr
b) 44 mole/hr
c) 70 mole/hr
d) 98.5 mole/hr

Answer: c [Reason:] 0.2 = – (-3.6)/(4 + 0.2*R), => R = 70 mole/hr.

10. What is the value of G?
a) 10 mole/hr
b) 30 mole/hr
c) 70 mole/hr
d) 80 mole/hr

Answer: d [Reason:] G = 10 + 70 = 80 mole/hr.

Answer question 11 – 15 for the following diagram. The reaction is C6H12 -> C6H6 + 3H2, the overall conversion of C6H12 is 80%, and single pass conversion rate is 20%. 11. What is the extent of reaction?
a) 4
b) 8
c) 12
d) 16

Answer: d [Reason:] Moles of C6H12 in feed = 80, => extent of reaction = (80*0.2 – 80)/(-1) = 64.

12. How many moles of benzene are formed?
a) 8
b) 16
c) 32
d) 64

Answer: d [Reason:] Moles of benzene formed = 64*1 = 64.

13. What is the value of P?
a) 64 mole/hr
b) 192 mole/hr
c) 272 mole/hr
d) 512 mole/hr

Answer: c [Reason:] Moles of benzene = 64, moles of hexane = 80 – 80*0.8 = 16, moles of H2 = 64*3 = 192, => P = 64 + 16 + 192 = 272 mole/hr.

14. What is the value of R?
a) 600 mole/hr
b) 720 mole/hr
c) 880 mole/hr
d) 960 mole/hr

Answer: a [Reason:] 0.2 = – (-64)/(80 + 0.4*R), => R = 600 mole/hr.

15. What is the value of G?
a) 600 mole/hr
b) 700 mole/hr
c) 800 mole/hr
d) 900 mole/hr

Answer: b [Reason:] G = 100 + 600 = 700 mole/hr.

## Set 3

1. How many recycle streams are there in the following process? a) 1
b) 2
c) 3
d) 4

Answer: b [Reason:] 2 recycle streams are there which goes from second reactor to first.

2. How many streams are there in the following process? a) 1
b) 2
c) 3
d) 4

Answer: c [Reason:] There are three recycle streams, from second to first, from third to first, from second to first.

3. How many recycle streams are there in the following process? a) 1
b) 2
c) 3
d) 4

Answer: b [Reason:] There are two recycle streams, one from top reactor to left most reactor and other from right most to left most.

4. What is the value of P, if the value of A = 10 mole, B = 15 mole, C = 6 mole, D = 8 mole? a) 2 mole
b) 5 mole
c) 6 mole
d) 8 mole

Answer: a [Reason:] Material balance for whole process, A = P + D, => P = 2 mole.

5. In the diagram below, what is the left most box? a) Mixer
b) Separator
c) Processor
d) None of the mentioned

Answer: a [Reason:] Any reactor in which recycled matter goes in is a mixer.

Answer question 6 – 10 for the following diagram. 6. What is the value of P?
a) 105.4 mole/hr
b) 254.3 mole/hr
c) 412.1 mole/hr
d) 520.8 mole/hr

Answer: d [Reason:] Overall NaOH balance, 1000*0.5 = P [0.9*1 + 0.1*0.6], => P = 520.8 mole/hr.

7. What is the value of W?
a) 227.5 mole/hr
b) 479.1 mole/hr
c) 568.2 mole/hr
d) 721.4 mole/hr

Answer: b [Reason:] Overall water balance, 1000*0.5 = 520.8*0.1*0.6 + W*1, => W = 479.1 mole/hr.

8. What is the value of C?
a) 104.1 mole/hr
b) 213.5 mole/hr
c) 342.2 mole/hr
d) 416.4 mole/hr

Answer: d [Reason:] First reactor balance, KOH: 1000*0.5 + C*0.6 = B*0.8, => 4B = 3C + 2500, H2O: 1000*0.5 + C*0.4 = 479.1*1 + B*0.2, => 209 + 4C = 2B, solving both equations we get C = 416.4 mole/hr.

9. What is the value of C?
a) 945.5 mole/hr
b) 1874.6 mole/hr
c) 2474.5 mole/hr
d) 3951.2 mole/hr

Answer: b [Reason:] First reactor balance, KOH: 1000*0.5 + C*0.6 = B*0.8, => 4B = 3C + 2500, => B = 1874.6 mole/hr.

10. What is the value of F for the B solution to remain 80% KOH if the recycle stream is removed?
a) 123.1 mole/hr
b) 666.5 mole/hr
c) 988.4 mole/hr
d) 1499.8 mole/hr

Answer: d [Reason:] Second reactor balance, KOH: B*0.8 = C*0.6 + 520.8*[0.9*1 + 0.1*0.6], => 4B = 3C + 2499.8, H2O: B*0.2 = C*0.4 + 520.8*[0.1*0.4], => B = 2C + 104.1, solving both equations we get B = 937.4 mole/hr. First reactor balance, KOH: F*0.5 = 937.4*0.8, => F = 1499.8 mole/hr.

Answer question 11 – 15 for the following diagram. 11. What is the value of P?
a) 14.4 mole/hr
b) 36.3 mole/hr
c) 48.1 mole/hr
d) 65.9 mole/hr

Answer: b [Reason:] Overall NaCl balance: 100*0.05 = B*0.12 + P*0.01, => 12B + P = 500, Overall water balance: 100*0.95 = B*0.88 + P*0.99, => 88B + 99P = 9500, => Solving both equations we get P = 36.3 mole/hr.

12. What is the value of B?
a) 18.9 mole/hr
b) 38.6 mole/hr
c) 45.2 mole/hr
d) 72.1 mole/hr

Answer: b [Reason:] Overall NaCl balance: 100*0.05 = B*0.12 + P*0.01, => 12B + P = 500, => B = 38.6 mole/hr.

13. What is the value of H?
a) 250 mole/hr
b) 400 mole/hr
c) 525 mole/hr
d) 750 mole/hr

Answer: a [Reason:] First reactor balance, NaCl: 100*0.05 + G*0.1 = H*0.08, => 8H = 10G + 500, H2O: 100*0.95 + G*0.9 = H*0.92, => 92H = 90G + 9500, Solving both equations we get H = 250 mole/hr.

14. What is the value of G?
a) 50 mole/hr
b) 100 mole/hr
c) 150 mole/hr
d) 200 mole/hr

Answer: c [Reason:] First reactor balance, NaCl: 100*0.05 + G*0.1 = H*0.08, => 8H = 10G + 500, => G = 150 mole/hr.

15. What is the value of C?
a) 126.1 mole/hr
b) 150.8 mole/hr
c) 178.4 mole/hr
d) 196.3 mole/hr

Answer: d [Reason:] Processor balance, NaCl: 0.08*250 = 0.01*36.3 + 0.1*C, => C = 196.3 mole/hr.

## Set 4

1. A vessel contains air and vapor H2O, if the partial pressure of vapor H2O is 4 atm and pressure of air is 1 atm, what is the fraction of vapor H2O in vessel?
a) 0.25
b) 0.5
c) 0.75
d) 0.8

Answer: c [Reason:] yH2O = 15/(15 + 5) = 0.75.

2. A container has vapor H2O with partial pressure 9 atm, what is fraction of air in the vessel?
a) 0.1
b) 0.4
c) 0.7
d) 0.9

Answer: a [Reason:] Pressure of air = 1 atm, => fraction of air = 1/(1 + 9) = 0.1.

3. A system has partial pressure 500 mm Hg and the saturated water vapor has partial pressure 25 mm Hg, what is the fraction of water vapor in the system?
a) 0.01
b) 0.05
c) 0.1
d) 0.5

Answer: b [Reason:] Fraction of water vapor = 25/500 = 0.05.

4. The partial pressure of a system at 49oC is 125 mm Hg and saturated ammonia vapor has partial pressure 20 mm Hg at 49oC, what is the fraction of ammonia vapor in the system?
a) 0.04
b) 0.08
c) 0.16
d) 0.24

Answer: [Reason:] Fraction of ammonia vapor = 20/125 = 0.16.

5. What is the dew point of water at standard conditions?
a) 32oF
b) 212oF
c) 372oF
d) 584oF

Answer: b [Reason:] Dew point of water at standard conditions is 100oC = 212oF.

6. CH4 is burnt with 100% excess O2 at atmospheric pressure, if CH4 conversion is 100%, what is the pressure at dew point?
a) 0.2 atm
b) 0.4 atm
c) 0.6 atm
d) 0.8 atm

Answer: b [Reason:] CH4 + 2O2 -> CO2 + 2H2O, Basis: 100 moles of CH4, => moles of O2 reacted = 200, => Product contains, moles of CO2 = 100, moles of H2O = 200, moles of O2 = 200. => Fraction of H2O = 0.4, => Pressure at due point = 1*0.4 = 0.4 atm.

7. Ethane is burnt with 20% excess O2 at 2 atm, conversion rate of ethane is 50%, what is the pressure at due point?
a) 0.45 atm
b) 0.75 atm
c) 0.9 atm
d) 1 atm

Answer: c [Reason:] C2H6 + 3.5O2 -> 2CO2 + 3H2O, Basis: 100 moles of ethane => moles of O2 reacted = 175, => Product contains, moles of CO2 = 100, moles of C2H6 = 50, moles of O2 = 35, moles of H2O = 150, => fraction of H2O = 0.45, => Pressure at due point = 2*0.45 = 0.9 atm.

8. Ethanol is burnt with required amount of O2 at atmospheric pressure, the conversion of ethanol is 100%, what is the pressure at due point?
a) 0.2 atm
b) 0.4 atm
c) 0.6 atm
d) 0.8 atm

Answer: c [Reason:] C2H5OH + 3O2 -> 2CO2 + 3H2O, Basis: 100 moles of ethanol, => moles of O2 reacted = 300, => Product contains, moles of CO2 = 200, H2O = 300, fraction of H2O = 0.6, => Pressure at due point = 0.6*1 = 0.6 atm.

9. 100 KPa of water vapor at 27oC is saturated with air in a system, what is the mole fraction of air?
a) 0.230
b) 0.306
c) 0.425
d) 0.503

Answer: d [Reason:] Fraction of air = 101.3/(100 + 101.3) = 0.503.

10. 10 KPa of water vapor at 27oC is saturated with air in a system, what is the mole fraction of air?
a) 0.21
b) 0.46
c) 0.79
d) 0.91

Answer: d [Reason:] Fraction of air = 101.3/(10 + 101.3) = 0.91.

## Set 5

1. What is the work done by an open steady-state system if heat transferred to it is 10 J?
a) 0
b) -10 J
c) 10 J
d) Cannot be determined

Answer: c [Reason:] In steady state system ∆E = 0, => Work done by the system = 10 J.

2. In a closed system, change in internal energy is 10 J and heat transferred to it is 10 J what is the work done by the system?
a) 0
b) 10 J
c) -20 J
d) 20 J

Answer: a [Reason:] ∆E = W + Q, => W = 10 – 10 = 0.

3. In an open system, what is the enthalpy change is heat taken from it is 10 J?
a) 0
b) -10 J
c) 10 J
d) Cannot be determined

Answer: b [Reason:] For an open system, ∆H = Q, => ∆H = -10 J.

4. In an open, steady state adiabatic process if the change in kinetic energy is 10 J and change in potential energy is 15 J, what is the work done by the system?
a) 10 J
b) 15 J
c) 25 J
d) -25 J

Answer: d [Reason:] -W = ∆KE + ∆PE = 10 + 15 = 25, => W = -25 J.

5. In an open, steady state system if the heat transferred is 10 J and work done by the system is 5 J, what is the change in Potential energy (neglect kinetic energy)?
a) 5 J
b) -5J
c) 15 J
d) -15 J

Answer: a [Reason:] W + Q = ∆PE, => ∆PE = 10 – 5 = 5 J.

6. Which of the following is true for isothermal ideal gases?
a) ∆T = 0
b) ∆U = 0
c) ∆H = 0
d) All of the mentioned

Answer: d [Reason:] For isothermal ideal gases, ∆T = 0, => ∆U = 0, => ∆H = 0.

7. If the system is insulated which of the following might the process be?
a) Isothermal
b) Isobaric
c) Isochoric

Answer: d [Reason:] Since the system is insulated, => Q = 0, => Process is adiabatic.

8. What is the work done for an isochoric process?
a) Positive
b) Negative
c) Zero
d) Cannot say

Answer: c [Reason:] W = -PdV, for isochoric process dV = 0, => W = 0.

9. What is the work done for ideal gas isothermal process?
a) Zero
b) Equal to heat transferred
c) Equal to change in internal energy
d) Cannot say

Answer: b [Reason:] For ideal gas isothermal process ∆U = 0, => Work done is equal to heat transferred.

10. What is the work done for an isobaric process?
a) P∆V
b) V∆P
c) PV
d) None of the mentioned

Answer: a [Reason:] For isobaric process ∆P = 0, => W = P∆V.

11. For an ideal gas isobaric process at pressure 1 Pa, the volume of the gas changes from 1 m3 to 5 m3, what is the work done by the gas?
a) 1 J
b) 4 J
c) 5 J
d) 10 J

Answer: b [Reason:] W = P∆V = 1*(5 – 1) = 4 J.

12. For an ideal gas isothermal process what is the work done by the gas from 1 m3 to 2 m3 at 300 K?
a) 0.69 kJ
b) 1.24 kJ
c) 1.73 kJ
d) 2.45 kJ

Answer: c [Reason:] W = RTln (V2/V1) = 8.314*300 ln (2/1) = 1.73 kJ.

13. For an ideal gas isothermal process what is the work done by the gas from 4 Pa to 1 Pa at 300 K?
a) 1.22 kJ
b) 2.15 kJ
c) 2.83 kJ
d) 3.45 kJ

Answer: d [Reason:] W = RTln (P1/P2) = 8.314*300 ln (4/1) = 3.45 kJ.

14. For an ideal gas isochoric process what is the work done by the gas from 2 Pa to 1 Pa?
a) 0
b) 1 J
c) 200 J
d) None of the mentioned