Constant temperature process 3

v

dT T

ds = −C D. Constant entropy process 4.

P

dT T

ds = −C

Codes: A B C D A B C D

(a) 3 2 1 4 (b) 2 4 3 1

(c) 3 4 1 2 (d) 1 3 4 2

Properties of Mixtures of Gases

IAS-21. If M1, M2, M3, be molecular weight of constituent gases and m1, m2, m3… their corresponding mass fractions, then what is the molecular weight M of the

mixture equal to? [IAS-2007]

(a)

m M

_{1 1}

+ m M

_{2 2}

+ m M

_{3 3}

+ ...

(b)

1 1 2 2 3 3

1

...

m M +m M +m M + (c)

1 1 2 2 3 3

1 1 1

...

m M +m M +m M + (d)

3

1 2

1 2 3

1

...

m

m m

M M M

⎛ ⎞

⎛ ⎞ ⎛ ⎞

+ + ⎜ ⎟ +

⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠

Properties of Gasses & Gas Mix.

S K Mondal’s Chapter 8

IAS-22. The entropy of a mixture of pure gases is the sum of the entropies of

constituents evaluated at [IAS-1998]

(a) Temperature and pressure for the mixture

(b) Temperature of the mixture and the partical pressure of the constituents (c) Temperature and volume of the mixture

(d) Pressure and volume of the mixture

Properties of Gasses & Gas Mix.

S K Mondal’s Chapter 8

Answers with Explanation (Objective) Previous 20-Years GATE Answers

GATE-1. (b). Let no of mole = n

mC 22.5 1.24 0.716

Δ = Δ Δ = Δ = = ∴ = + =

Properties of Gasses & Gas Mix.

S K Mondal’s Chapter 8

⎛ ⎞

= ⎜ ⎟

⎝ ⎠

⎛ ⎞

= × × ⎜⎝ ⎟⎠=

2 1 1

1

ln V V

0.030

800 0.015 ln 8.32kJ/Kg 0.015

PV

GATE-7. Ans. (b) Remember if we mix 2 mole of oxygen with another 2 mole of other gas the volume will be doubled for first and second constituents ln 2Rln2

V nR V S

initial total =

=

Δ

∴

Total Entropy change = 4Rln2 So, Entropy change per mole=Rln2. And it is due to diffusion of one gas into another.

Previous 20-Years IES Answers

IES-1. Ans. (a) Both A and R correct and R is the correct explanation of A

IES-2. Ans. (b) For perfect gas, both the assertion A and reason R are true. However R is not the explanation for A. A provides definition of perfect gas. R provides further relationship for enthalpy and internal energy but can't be reason for definition of perfect gas.

IES-3. Ans. (b) As internal energy is a function of temperature only. In isothermal expansion process no temperature change therefore no internal energy change. A Reversible isothermal expansion process is constant internal energy process i.e. dU = 0

( )

= +

∴ = =

∴ =

∵

∵ dQ dU dW dQ dW dU 0

Work done during the process 100kW IES-4. Ans. (a)

IES-5. Ans. (d) (a) True. A water film, if formed, will act as a very poor conductor of heat and will not easily let the heat of the furnace pass into the boiler. An oil film if present, is even worse than water film and the formation of such films inside the boiler must be avoided.

(b) Since the mass and material are the same, the volumes must also be the same. For the same volume, the surface area of the plate is the greatest and that of the sphere is the least. The rate of loss of heat by radiation being proportional to the surface area, the plate cools the fastest and the sphere the slowest.

(c) True, for a monoatomic gas, C = ₁ ³

2 R and for a diatomic gas, C = ₁ ⁵ 2 R.

Since the mixture has two moles, the value of C for the mixture = ₁ 1 3R 5R

2 2 2

⎛ + ⎞

⎜ ⎟

⎝ ⎠= 2 R (d) False, The average kinetic energy of 1 g of an ideal gas = ³

2 RT

M

Where M is the molecular weight of the gas and it is different gases, as the value of M will be different.

IES-6. Ans (b) In Perfect gas intermolecular attraction is zero. It will be only possible when intermolecular distance will be too high. High temperature or low pressure or both cause high intermolecular distance so choice 1 and 3.

IES-7. Ans. (a) The correct sequence for decreasing order of the value of characteristic gas constants is hydrogen, nitrogen, air and carbon dioxide.

IES-8. Ans. (b)

Properties of Gasses & Gas Mix.

S K Mondal’s Chapter 8

IES-9. Ans. (d) we know that P =

1

²

3 ρ C

If momentum is zero then C must be zero. Hence P would be zero. That will occur at absolute zero temperature. But note here choice (a) has

in defined temp. –273K which is imaginary temp.

IES-10. Ans. (a) IES-11. Ans. (c)

IES-12. Ans. (b) Joule’s law states that for an Ideal gas internal energy is a function of temperature only. u = ƒ(T). But this is not Ideal gas it is real gas.

IES-13. Ans. (a) We know that at critical point a = 3PcVc2 ; b = Vc/3 and R =

Tc PcVc

3 8

IES-14. Ans. (a) 1 is false. At very low pressure, all the gases shown have z

≈

1 and behave nearly perfectly. At high pressure all the gases have z > 1, signifying that they are more difficult to compress than a perfect gas (for a given molar volume, the product pv is greater than RT). Repulsive forces are now dominant. At intermediate pressure, must gasses have Z < 1, including that the attractive forces are dominant and favour compression.

IES-15. Ans. (a) IES-16. Ans. (d) IES-17. Ans. (c) IES-18. Ans. (d) IES-19. Ans. (d)

IES-20. Ans. (b) During adiabatic process, work done = change in internal energy.

Since control man (so case of closed system). Intercept of path on X-axis is the work done by the process.

W = area of ∆A12 + area of □A2CB

W = ¹

2 х (3 - 1) х 200 + 100 х (3 - 1) = 200 + 200 = 400 kJ

W = ¹

2 (300 - 100) х 2 + 100 х 2 = 200 + 200 = 400 kJ

∴ From 1 →2.

U + Q = 1 U + W. ₂

U – 1 U = W – Q = 400 – 200 = 200 kJ. ₂ From 2 →1

Work done will be same Since adiabatic So Q = 0

U + Q = 2 U + W ₁

W = U – ₂ U = – (₁ U – ₁ U ) = – 200 kJ. ₂

IES-21. Ans. (a) For isothermal process, _{1 1 2 2}

,

_{1 1} ¹

0.55,

₁

0.055

³

10

p v = p v or p v = p × v = m

For adiabatic process

^{p v}

^1.4

^{= p v}

^1.4

^{, or p} ( ^0.055 )

^1.4

^{= p}

¹

^{× v}

^1.4

^{or v = 0.055 10}

^1.4

^{= 0.45 m}

³

Properties of Gasses & Gas Mix.

S K Mondal’s Chapter 8

IES-22. Ans. (b) Work done by the gas during filling process = −

∫

^vdp

⁼

(

⁴₃^π13

) ⁽

¹⁰⁰

⁾

^{= 418.9 kJ}

IES-23. Ans. (c) Turbine work = area under curve R–S =

∫

^{P dv}

( )

( )

= × − +

= × − + =

3 5

1 bar 0.2 0.1 m 1000 Nm

10 0.2 0.1 Nm 1000Nm 11000Nm IES-24. Ans. (b)

IES-25. Ans. (a) PV = constant, C

⇒ log P + log V = log C m = -1 1

Pv = C 4

⇒ log P + q log V = log C m = -q = - 1.4 2

∴ m > ₂ m ₁ IES-26. Ans. (d)

IES-27. Ans. (d) Since the temperature remains constant, the process is isothemal.

∴ Work-done in the process, W = 2.303 nRT log ²

1

V V

⎛ ⎞

⎜ ⎟

⎝ ⎠

= 2.303 х 3 х 8.315 х 8.315 х 300 log 1

2

⎛ ⎞⎜ ⎟

⎝ ⎠

= – 5188 J.

The negative sign indicates that work is done on the gas.

IES-28. Ans. (b) Tds =du pdv+

⇒ +

⇒ +

⇒ +

V V

V

Τds = C dT pdv dT P

ds = C dv

T T

dT R

ds = C dv

T V

Integrating the above expression

⇒ ₂ − ₁ = _V ² + ²

1 1

T V

S S C In RIn

T V

For isothermal process undergone by ideal gas.

( )

⇒ ₂− ₁ = _P − _V ²

1

S S C C InV V IES-29. Ans. (a)

IES-30. Ans. (a) IES-31. Ans. (c)

IES-32. Ans. (c) ^{dQ du pd= + υ υ+ pd−υdp d u p=}

(

^{+ υ}

)

^{−υdp dh= −υdp}

( ) ( )

_{p p}

if dp 0 or p const. these for dQ= = = dh

Properties of Gasses & Gas Mix.

S K Mondal’s Chapter 8

IES-33. Ans. (a)

IES-34. Ans. (d) By compressing a saturated vapour, its vapours condense and pressure remains unchanged. Remember it is not gas.

IES-35. Ans. (b)

( )

= ⇒ =

+

1 2 1 1

1 2 2

V V V 3V

T T 273 27 T

⇒ T₂ =300× 3 = 900 = 627°K C

IES-36. Ans. (a) Constant volume (isochoric) process: An example of this process is the heating or cooling of a gas stored in a rigid cylinder. Since the volume of the gas does not change, no external work is done, and work transferred ΔW is zero. Therefore from 1st law of thermodynamics for a constant volume process:

1 2 2

1 2 2 1

1

W 0

Q dU U U

=

=

∫

= −

IES-37. (c) In a pressure cooker, the volume of the cooker is fixed so constant volume process but for safety some of steam goes out to maintain a maximum pressure. But it occurs after proper steaming.

IES-38. Ans. (b) IES-39. Ans. (c) IES-40. Ans. (a) IES-41. Ans. (b) IES-42. Ans. (d) IES-43. Ans. (c)

IES-44. Ans. (c) Perfect gas

T = 27 + 273 = 300 k 1

V = initial volume 1

V = final volume,2 V = ₂ 2V ₁ T = ? 2

(i) Constant Pressure Process V Tα

Properties of Gasses & Gas Mix.

S K Mondal’s Chapter 8

1 2 (iii) Isentropic Process

r 1 r 1

Previous 20-Years IAS Answers

IAS-1. Ans. (b) Boyle’s law: It states that volume of a given mass of a perfect gas varies inversely as the absolute pressure when temperature is constant.

IAS-2. Ans. (a)

IAS-5. Ans. (a) Apply equation of states

( )

IAS-6. Ans. (c ) Apply equation of states

1 which is depicted in figure (c).

IAS-8. Ans. (d) a = 3 pc Vc2, b = 8

IAS-9. Ans. (b) According to dimensional homogeneity law unit of molar-volume and ‘b’ must be same. i.e. m³/mole

IAS-10. Ans. (b) IAS-11. Ans. (c)

IAS-12. Ans. (d) Van der Waals equation

( )

Properties of Gasses & Gas Mix.

S K Mondal’s Chapter 8

At critical point a= 3pcVc2, b= 8

IAS-13. Ans. (a) In adiabatic mixing there is always increase in entropy so large amount of irreversibility is these.

IAS-14. Ans. (c)

IAS-15. Ans. (d) For reversible isothermal expansion heat supplied is equal to work done during the process and equal to Q = W = mRT1 2

∵ Temperature constant so no change in internal energy dQ = dU + dW; dU = 0 Therefore dQ = dW.

IAS-16. Ans. (d) In reversible isothermal process temperature constant. No change in internal energy. So internal energy constant

dQ = δ u + δ W as δ u = 0, dQ = dW

IAS-17. Ans. (d) IAS-18. Ans. (d)

IAS-19. Ans. (b) Work requirement 1. Isothermal – area under

In document thermo (Page 113-121)