what is this mean line H, 1 mol 0, 2 mol Adiabat 63] alculate the entropy change

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what is this mean

line

H, 1 mol 0, 2 mol Adiabat 63] alculate the entropy change if we remove the adiabatic plate as following Fi Gem for O2 and N2 gases are the same : Cprm-29.0J/K.mole tic plate as following Figure. Asoute 66] An ice block having a Adiabatic plate vessel containing 100 185 KJ/Kg-K over t (1)Calculate the final (2)Calculate the entro (1)T 10x 334+10 RI T=38.2 e2): ASt = Adiabatic box 1 mol O 2 mol N2 290K 310K (2)(29.0)(310-T-(1)(29.0)(7-290): T= 303.3 K 303.3 310 AS 303.3 [64]A ,,entropy (Apd (B)-O (C)-0 (D) 67] 1.388 mol sam essure (1 bar)i mol. C of wa S? [65]E(3 8 10 5 2

Explanation / Answer

The calculation for entropy change has been displayed.

We know that,

Q = n Cv (T2 - T1)

Using the abive relation, they have calculated the common temperature.

Q(lost) = Q(Gain)

Since second part is at high temperature it will loose heat energy

2 x 29 x (310 - T) = 1 x 29 (T - 290)

T = 303.3 K

We know that entropy change is given by:

del-S = n Cv ln (T/T0) + R ln V2/V1

Sice, V = n R T/P ; P is same for 2 so

V is prop to number of moles. So

for the given system it will be:

delta-S = S1 + S2 + R [n2 ln (n2/3) + n1 ln(n1)/3)]

deltaS = 2 x 29 ln(303.3/310) + 1 x 29 ln(303.3/290) - 8.3145 [2 x ln 2/3 + 1 x ln 1/3] = 15.9 kJ/K

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