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A 1.5-m 3 insulated rigid tank contains 2.7 kg of carbon dioxide at 100 kPa. Now paddle-wheel work is done on the system until the pressure in the tank rises to 150 kPa. Determine the entropy change of carbon dioxide during this process. Assume constant specific heats

Sagot :

Answer:

The entropy change of carbon dioxide = 0.719 kJ/k

Explanation:

Given:

1.5 m - 3 insulated rigid tank contains 2.7 kg of carbon dioxide at 100 kPa

The objective is to determine the entropy change of carbon dioxide

Formula used:

ΔS=

Solution:

On considering,

[tex]C_{P} =0.846 kJ/kg K\\C_V=0.657 kJ/kg k\\[/tex]

ΔS=[tex]mc_{v} lu\frac{p_{2} }{P_{1} }[/tex]

On substituting the values,

ΔS=[tex]2.7*0.657lu\frac{150}{100}[/tex]

ΔS=0.719 kJ/k

Cricetus

The entropy change is "0.719 kJ/K".

Given values are:

Mass of tank,

  • m = 2.7 kg

Pressure,

  • P₁ = 100 kPa

Rised pressure,

  • P₂ = 150 kPa

Assumption of constant specific heat is,

  • [tex]C_v = 0.657 \ kJ/kgK[/tex]

As we know the formula,

→ [tex]\Delta S = mC_v \ ln(\frac{P_2}{P_1} )[/tex]

         [tex]= (2.7)(0.657) \ ln (\frac{150}{100} )[/tex]

         [tex]= 1.7739\times 0.4055[/tex]

         [tex]= 0.7193 \ kJ/K[/tex]

Thus above answer is right.

Learn more:

https://brainly.com/question/20340354

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