Question

Derivation for workdone in an adiabatic process

Answer 1

First of all, let’s define terms- an adiabatic process is one in which the system is fully incsulated from the outside world. That is, there is no heat transfer from the system to the universe.

Remember that the first law of thermodynamics states (Q+W+u^1)=(u^2)(Q+W+u^1)=(u^2)

Where u^u^ is internal energy.

By definition of an adiabatic process, the only way that internal energy can change is via work, which is defined as the integral of P dV

Before we can solve this integral, we need to formulate a relationship between pressure and volume under the condition of an adiabatic process. Consider then, the differential form of the first law of thermodynamics (done by considering the limit as the increment of heat and work added to the gas goes to zero, and allowable under the assumption of reversibility), namely

dQ=dW+dU=0=PdV+dUdQ=dW+dU=0=PdV+dU

For an ideal gas, another equation of note is

dU=nCvdTdU=nCvdT

from the definition of Cv=dUdTCv=dUdT

Utilizing the ideal gas equation, we see that

nCvdT=nCvnR(PdV+VdP)nCvdT=nCvnR(PdV+VdP)

Once again substituting into the differential form, we see that

Cv+RRPdV+CvRVdP=0Cv+RRPdV+CvRVdP=0

Which, upon relatively simple integration, results in the equation

PVγ=P0Vγ0PVγ=P0V0γ

and allows integration to find that

P0Vγ0γ−1(V1−γ0−V1−γ1)P0V0γγ−1(V01−γ−V11−γ)

So, knowing the initial and final states of the gas results in a unique solution for work in an adiabatic process. Let me know if any steps were unclear!

Answer 2

${\huge{\underline{\underline{\mathfrak{\red{Answer:-}}}}}}$

An adiabatic process is one in which no heat is gained or lost by the system. The first law of thermodynamics with Q=0 shows that all the change in internal energy is in the form of work done. ... This condition can be used to derive the expression for the work done during an adiabatic process.