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A piston-cylinder device initially at 0.45-m3 contains nitrogen gas at 600 kPa and 300 K. Now the gas is compressed isothermally to a volume of 0.2 m3. The work done on the gas during this compression process is _____ kJ.

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Answer:

219kJ

Step-by-step explanation:

The work done (W) on a gas in an isothermal process is given by;

W = -P₁V₁ ln
(V_(2))/(V_1) -----------------(i)

Where;

P₁ = initial pressure of the gas

V₁ = initial volume of the gas

V₂ = final volume of the gas

From the question;

P₁ = 600kPa = 6 x 10⁵Pa

V₁ = 0.45m³

V₂ = 0.2m³

Substitute these values into equation (i) as follows;

W = -6 x 10⁵ x 0.45 x ln
(0.2)/(0.45)

W = -6 x 10⁵ x 0.45 x ln (0.444)

W = -6 x 10⁵ x 0.45 x -0.811

W = 2.19 x 10⁵

W = 219 x 10³

W = 219kJ

Therefore, the work done on the gas during the compression process is 219kJ

User Manthan Jamdagni
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