A 2-m3insulated rigid tank contains 3.2 kg of carbon dioxide at 120 kPa. Paddle-wheel work is done on the system until the pressure in the tank rises to 180kPa. Determine the entropy change in the carbon dioxide during this process. Assumeconstant specific heat and room temperature at 300 K

The change in entropy is found to be 0.85244 KJ/k

Explanation:

In order to solve this question, we first need to find the ration of temperature for both state 1 and state 2. For that, we can use Charles' law. Because the volume of the tank is constant.

P1/T1 = P2/T2

T2/T1 = P2/P1

T2/T1 = 180 KPa/120KPa

T2/T1 = 1.5

Now, the change in entropy is given as:

ΔS = m(s2 - s1)

where,

s2 = Cv ln(T2/T1)

s1 = R ln(V2/V1)

ΔS = change in entropy

m = mass of CO2 = 3.2 kg

Therefore,

ΔS = m[Cv ln(T2/T1) - R ln(V2/V1)]

Since, V1 = V2, therefore,

ΔS = mCv ln(T2/T1)

Cv at 300 k for carbondioxide is 0.657 KJ/Kg.K

Therefore,

ΔS = (3.2 kg)(0.657 KJ/kg.k) ln(1.5)

ΔS = 0.85244 KJ/k

answer:

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explanation:

answer:

1.3 + 20

explanation:

20.1.3