Two equilibrium states, a and b , of a fixed amount of monatomic ideal gas are l

Question

Two equilibrium states, a and b, of a fixed amount of monatomic ideal gas are linked by a two step process, a-->c-->b. You may assume cv=3/2R. For state a, P=103Pa, V=2,0m3, and T=300K. For state b, P=104Pa, V=1m3. The process a-->c is a quasistatic adiabatic process, and the process c-->b is a constant volume stirring process, which is adiabatic but not quasistatic.

a) calculate the temperature for the states b and c. Hence, calculate the increase in internal energy, U, of the gas for the two processes.

b) calculate the work done in the process a-->c using

Wac=(PaVka)/(k-1) ((1/Vck-1)-(1/Vak-1)) where k=cp/cv

c) the process c-->b is an adiabatic process, but the above equation cannot calculate the work performed. Why?

Explanation / Answer

(Pa, Va, Ta)=(103 Pas, 2 m3, 300 K)

(Pb, Vb, Tb)=(104 Pas, 1 m3, Tb)

(Pc, Vc, Tc)=(Pc, 1 m3, Tc)

Note that we have taken Vb=Vc=1m3 because c->b is constant volume process

Cv=3/2R, For ideal gas Cp=Cv+R=5/2R . So k=Cp/Cv=5/3

Also by ideal gas equation PV=nRT where R is the gas constant

Putting value of Pa, Va, Ta we can find that:

nR=103*2/300 = 0.687 J/K

Part a:

Since process a->c is adiabatic, we have

Ta*(Va)^(k-1) = Tc*(Vc)^(k-1)

This equation will give Tc = 476.22 K   

Also by ideal gas law,

Pa*Va/Ta = Pb*Vb/Tb

This eqn will give Tb = 151.46 K

Change in internal energy in process a->c = Cv*n*(Tc-Ta) =3/2 * n*R*(Tc-Ta) = 181.59 J

Similarly for process c->b

Change in internal energy in process c->b = Cv*n*(Tb-Tc) =3/2 * n*R*(Tb-Tc) = -334.67 J

Part b:

Applying ideal gas law between points c and a, we get

Pa*Va/Ta = Pc*Vc/Tc

From here we calculate Pc = 327.004 Pas

Formula for work done in adiabatic process can be written as Wa->c = (Pc*Vc-Pa*Va)/(1-k)

Putting all the values we get,

Wa->c = -181.51 J  

Part c:

We cannot use this equation for work done in the process c->b because this process is not quasistatic or reversible.

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