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3 years ago

Problem 1

Engineering

1 answer:

Elis [28]3 years ago

6 0

The work done on the gas during this process if the final pressure is 456 kPa is; -82.8 kJ

<h3>Workdone in Thermodynamics</h3>

We are given;

The initial pressure; P₁ = 96 kPa

Final Pressure; P₂ = 456 kPa

The gas is compressed according to;

p = aV + b

where;

a = -1200 kPa/m³

b = 600 kPa

Thus, at initial pressure P₁ = 96 kPa;

96 = -1200V₁ + 600

1200V₁ = 600 - 95

1200V₁ = 505

V₁ = 505/1200

V₁ = 0.42 m³

At Final Pressure P₂ = 456 kPa;

456 = -1200V₂ + 600

1200V₂ = 600 - 456

1200V₂ = 144

V₂ = 144/1200

V₂ = 0.12 m³

Formula for the workdone during the process is;

W_out = ¹/₂(P₁ + P₂)(V₂ - V₁)

W_out = ¹/₂(96 + 456)(0.12 - 0.42)

W_out = -82.8 kJ

Read more about workdone in thermodynamics at; brainly.com/question/12641937

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Explanation:

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