Thermodynamics Flashcards

Study and learn Thermodynamics hard q and imp concept (46 cards)

1
Q

What is thermal equilibrium?

A

Two systems are in thermal equilibrium with each other if they have the same temperature

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2
Q

Zeroeth Law of thermodynamics

A

If two systems A and B are separately in thermal equilibrium with a third system C, then A and B are also in thermal equilibrium with each other.

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3
Q

Internal Energy

A

Internal energy of a system is the sum of molecular kinetic and potential energies in the frame of reference to which the center of mass of the system is at rest

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4
Q

Intermolecular potential energy corresponds to

A

Volume

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5
Q

Intermolecular kinetic energy corresponds to

A

Temperature

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6
Q

Heat

A

Mode of energy transfer due to temperature difference between system and surroundings

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7
Q

Work

A

Mode of energy transfer brought about by means that do not involve temperature difference

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8
Q

Sign Convention

(Heat, Work, Internal Energy)

A

Heat Absorbed = +
Work Done By system = +
Increase in internal energy = +

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9
Q

Work Done

Derivation

A

dW = Fdx = PA dx = Pdv
W = integral (PdV)

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10
Q

Non-cyclic process

A

System does not return to its initial state

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11
Q

Work Done (Cyclic Process)

Explain what is + and -

A

Work taken in clockwise = +
Work taken in anti-clockwise = -

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12
Q

Cyclic Process

A

Any process in which the system returns to its initial state after undergoing a series of changes

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13
Q

First Law of Thermodynamics

A

If some heat is supplied to a system which is capable of doing work, then the quantity of heat absorbed by the system will be equal to the sum of increase in its internal energy and the external work done by system

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14
Q

First Law of Therm (Formula)

A

delta Q = delta U + delta W {or P delta V}

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15
Q

Specific heat during adiabatic change

A

Zero

delta Q = 0

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16
Q

specific heat during isothermal

A

Infinite

delta T = 0; 1/0 = infinity

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17
Q

Change in temp (on specific heat)

A

delta T = + specific heat = +
delta T = - specific heat = -

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18
Q

Why is Cp > Cv

specific heat at constant p > constant v

A
  • Heat at constant volume uses all energy to increase temp
  • Heat at constant p causes expansion = work against external pressure. Hence, additional amount of heat equivalent to work done is used
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19
Q

Cp - Cv formula

A

Cp - Cv = R / J

R is the gas constant

20
Q

Isothermal Process

A

Thermodynamic process at constant temp

21
Q

Isobaric process

A

Thermodynamic process at constant pressure

22
Q

Isochoric process

A

Thermodynamic process at constant volume

23
Q

Adiabatic

A

Thermodynamic process having no heat exchange

24
Q

Conditions for isothermal process

A
  1. Walls must be perfectly conducting to allow free exchange of heat
  2. Process of compression or expansion should be very slow
25
Isothermal Process (formula) | For P1, V1
P1V1 = P2V2
26
Explain effect of gas expanding isothermally
* Internal energy dependent only on temp --> delta U = 0 * delta Q = P delta V * If V increases = Q increases (+) * An amount of heat = **work done by gas** has to be supplied
27
Derive expression for work done during isothermal expansion
W = 2.303 nRT log V2 / V1 | actually derive it
28
Conditions for adiabatic
1. Walls of container must be perfectly insulated 2. compession or expansion should be sudden so heat is not exchanged
29
Adiabatic relation between P and V
PV^(Cp / Cv) = K | derive it
30
Adiabatic relation between P and T
[P^(1-Cp/Cv)] * T^(Cp/Cv) = constant | derive
31
Adiabatic relation between T and V
TV^(Cp/Cv - 1) = constant | derive itx
32
Work done in adiabatic
W = [1/(gamma - 1)] (P1V1 - P2V2) | derive
33
Work done in adiabatic (in moles)
W = nR/(gamma - 1) * (T1 - T2) | derive
34
First Law applied to isochoric
Q = delta U + delta W = delta U ( volume is constant) Q = nCv delta T
35
First Law applied to cyclic
Q = delta U + W = 0 + W = W (No change in internal energy) | Total heat absorbed = work done by system
36
First Law of therm in boiling
Uf - Ui = mL - P(Vf - Vi)
37
First Law applied to isobaric
W = nR(T2 - T1) Absorbed heat goes partly to increase internal energy
38
Second Law of Thermodynamics
A simple statement of the law is that heat always flows spontaneously from hotter to colder regions of matter. "Not all heat can be converted into work in a cyclic process."
39
Reversible Processes
Any process which can be made to proceeed in the reverse direction by variation in its conditions such that any change occuring in any part of reversible process
40
Conditions for reversible process
An infinitesimally slow compression and expansion of an ideal gas at constant temperature
41
Irreversible process
Any process which cannot be retaced in the reverse direction exactly
42
1g of water at 100∘C is converted into steam at the same temperature. If the volume of steam is 1671cm3, find the change in the internal energy of the system. Latent of steam = 2256 Jg−1. Given 1 atmospheric pressure = 1.013 × 10^5 Nm−2
2086.829J
43
Why is slope of adiabatic steeper than isothermal?
Adiabatic --> PV^gamma Isothermal --> PV Since gamma > 1; pressure inversely prop. (adiabatic > isothermal)
44
Quasi Static Process
State ariables of a thermodynamical system change infinitely slowly Hypothetical and reersible
45
Cyclic process properties
1. Isothermal 2. Delta U = 0
46
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