siddharth

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100 g of N2 at 300K were held by a piston at 30 atm. Pressure was released suddenly to become 10 atm, adiabatically. Calculate [tex] \delta S [/tex]. Cv=20.8 J/(K*mol)

Since this is adiabatic , q=0. So,

[tex] \delta U = w [/tex]

Now, as the pressure was released suddenly to become 10 atm, this must be an irreversible adiabatic expansion.

Also, from the statement, what I understand is that the external pressure was rapidly changed to 10 atm and the piston was allowed to expand till the pressure of the gas inside was also 10atm.

So the work done in this case must be

[tex] -10(V_2 - V_1) [/tex]

Also,

[tex] V_1 = \frac{nRT_1}{P_1} = 2.93 L [/tex]

So,

[tex] (3.57)(20.8)(T_2-300)=-10(V_2 - 2.93)*10^2 [/tex]

and

[tex] P_1V_1/T_1 = P_2V_2/T_2 [/tex]

=> [tex] (30)(2.93)/300 = (10)(V_2)/(T_2) [/tex]

Solving these two equations, I get

[itex] T_2 = 243.44 K [/itex] and [itex] V_2 = 7.13 L [/itex]

As Entropy is a state function, considering the reversible path,

[tex] \delta S = nCvln(T_2/T_1) + nRln(V2/V1) [/tex]

[tex] \delta S = 0.854 [/tex]

The answer which is given is -0.692

Can you check my work and tell me if I made any calculation or conceptual mistakes?