Isothermal and adiabatic process Flashcards
What is a Diathermal wall.
A wall that allows the flow of heat. Two systems separated by diathermal walls are said to be in thermal contact.
Define Adiathermal.
Without the flow of heat. A system bounded by adiathermal walls is thermally isolated. Any work done on such a system produces adiathermal change.
Define Adiabatic.
adiathermal (without the flow of heat) and reversible.
Define a Reversible Process.
In thermodynamics, a reversible process is a process whose direction can be returned to its original position by inducing infinitesimal changes to some property of the system (such as P, V, … equilibrium states) via its surroundings. Throughout the entire reversible process, the system is in thermodynamic equilibrium with its surroundings.
Blundell & Blundell defn.: sufficiently slow process so that the gas remains in equilibrium throughout the entire process and passes seamlessly from one equilibrium state to the next, each equilibrium state differing from the previous one by an infinitesimal change in the system parameters.
Define Quasistatic Process.
A thermodynamic process that happens slowly enough for the system to remain in internal equilibrium.
A quasi-static process refers to an idealized or imagined process where the change in state is made infinitesimally slowly so that at each instant, the system can be assumed to be at a thermodynamic equilibrium with itself and with the environment.
Ex. Hair growing, plant growing, nails growing, galaxy expanding, expansion of railroad tracks in the daytime.
Reversibility requires the absence of __________, not the generation of _________.
friction; heat.
Any reversible process is a ____________ one but not any __________ process is a reversible one.
quasistatic.
Calculate the heat change in a reversible isothermal expansion of an ideal gas.
ΔQ = RT·ln(V2/V1)
Here, the internal energy stayed the same, volume increase ⇒ energy density has gone down along with pressure due to p = 2/3·u.
Define Isothermal.
Constant temperature.
This expansion/compression concludes that the internal energy does not change since work gained/lost is equal to the the heat lost/gained.
dW = -dQ
Isothermal expansion/compression.
ΔT = 0.
Define a irreversible process.
A non-zero change (rather than a sequence of infinitesimal changes) is made to the system, and therefore the system is not in equilibrium throughout the process.
Which type of processes often occur and what is the reason?
Irreversible processes.
Reason: there are lots of more ways that the energy can be distributed in heat than in any other way, and this is therefore the most probable.
Define Adiathermal.
Without the flow of heat.
A change is ___________ if it is both adiathermal & reversible.
adiabatic.
A system is bounded by adiathermal walls is said to be ________________.
thermally isolated.
Assume an adiabatic expansion for 1 mole of ideal gas. Derive TVγ-1 = C and PVγ = D, where C & D are constants.
Define Isotherms and Adiabats.
Isotherms: Lines of constant temperature, as would be followed in an isothermal expansion.
Adiabats: Lines followed by an adiabatic expansion in which heat cannot enter or leave the system.
Is work a state function?
No, since it is not an exact differential. Work depends on the path in which it is applied. State functions are path independent thus are exact differentials.
Define the work done in a reversibe process.
dW = -pdV,
Think of the frictionless piston.
Define the work done by a irreversible process.
W = F · Δx
Dervive the result dW = -dQ of an isothermal expansion/compression.
dU = (∂U/∂T)V dT + (∂U/∂V)T dV
Assuming ET = KE, U(N,T) = 3/2 NkBT ⇒ dV = 0.
Isothermal Expansion/Compression ⇒ ΔT = 0 ⇒ dU = 0.
By the 1st law of thermodynamics, dW = -dQ.
How does adding heat change the internal energy of gas? Furthermore, how is heat capacity related this?
What is the difference between equilibrium and an steady state.
In terms of temperature,
two objects are in thermal equilibrium when they have the same temperature.
A system is in a steady state when the (dT/dt) = 0, in other words temperature is constant with respect with time (this is accomplished when (dQ/dt)in = (dQ/dt)out of the system.
steady state is not an equilibrium state.
