An insulated box is separated by insulated partition into two parts, which is oc

Question

An insulated box is separated by insulated partition into two parts, which is occupied by Ar(ideal gas,1mol,300K,100kPa) and N2 ( ideal gas, 2mol 400K, 200kPa)

b) Find the S of mixing after the partition is removed. [Hint: The mixing process can be divided to two parts: constant-volume process and isothermal process.]

I had tried to answer this qustion, but i did not make sure that is correct. If the answer was wrong, help me to solve.

S=S(Ar)+S(N2)

S(Ar)=nCvm In(T2/T1)+nR In(P1/P2)

S(Ar)=(1)(1.5)(8.314)(In100/300)+(1)(8.314)(In100/300)

S(Ar)=-22.83 JK-1

S(N2)=(2)(2.5)(8.314)(In100/400)+(2)(8.314)(In200/300)

S(N2)=-64.37 JK-1

S=-22.83-64.37=-87.20JK-1

Explanation / Answer

Entropy for the system after mixing,

dS = dS(Ar) + dS(N2)

dS(Ar) = nCvln(T2/T1) + nRln(P1/P2)

           = 1 x 1.5 x 8.314 ln(400/300) + 1 x 8.314 ln(100 x 10^3/200 x 10^3)

           = 5.98 - 5.76

           = 0.22 J/K

dS(N2) = nCvln(T2/T1) + nRln(P1/P2)

           = 2 x 2.5 x 8.314 ln(300/400) + 2 x 8.314 ln(200/100)

           = -11.96 + 11.53

           = -0.43 J/K

dS = 0.22 - 0.43 = -0.21 J/K

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