Question

N atoms of an ideal gas are contained in a cylinder with insulating (adiabatic) walls, closed at one end by a piston.The initial volume is Vi and the initial temperature is Ti . Find the change in temperature, pressure and entropy that would occur if the volume were suddenly increased to Vf by withdrawing the piston (Vf > Vi).

Answer 1

When an ideal gas is compressed adiabatically (Q=0), work is done on it and its temperature increases; in an adiabatic expansion, the gas does work and its temperature drops. Adiabatic compressions actually occur in the cylinders of a car, where the compressions of the gas-air mixture take place so quickly that there is no time for the mixture to exchange heat with its environment. Nevertheless, because work is done on the mixture during the compression, its temperature does rise significantly.

Another interesting adiabatic process is the free expansion of a gas. Figure shows a gas confined by a membrane to one side of a two-compartment, thermally insulated container. When the membrane is punctured, gas rushes into the empty side of the container, thereby expanding freely. Because the gas expands “against a vacuum” (p=0)(p=0), it does no work, and because the vessel is thermally insulated, the expansion is adiabatic. With Q=0 and W=0 in the first law, ΔEint=0, so Einti=Eintf for the free expansion.

Here the gas is ideal, the internal energy depends only on the temperature. Therefore, when an ideal gas expands freely, its temperature does not change.

When the gas expands by dV, the change in its temperature is dT. The work done by the gas in the expansion is dW=pdV;

dQ=0dW=pdV;

dQ=0

because the cylinder is insulated; and the change in the internal energy of the gas is, dEint=CVdT.

Therefore, from the first law,

CVdT=0−pdV=−pdV,

so

dT=−pdV/CV.

When the gas expands adiabatically and in the insulated vessel.

Also, for 1 mol of an ideal gas,

d(pV)=d(RT),

so

pdV+Vdp=RdT

and

dT=pdV+VdpR.

We now have two equations for dT. Upon equating them, we find that

CvVdp+(Cv+R)pdV=0.

Now, we divide this equation by pV and use Cp=CV+R. We are then left with

Cv(dp/p)+Cp(dV/V)=0,

which becomes

(dp/p)+γ(dV/V)=0.

Now coming to our point directly,

Because we are modeling the process as a adiabatic compression of an ideal gas, we have pVγ=constantp and pV=nRT. The work needed can then be evaluated with W=∫_v1^v2 pdV.

For an adiabatic compression we have after the compression, the pressure of the mixture is

pf=p1(Vi/Vf)γ.

from the ideal gas law, the temperature of the mixture after the compression is

Tf=(Pf Vf/ Pi Vi)Ti.

The work done by the mixture during the compression is

W=∫_Vi^Vf pdV.