A 0.8-m3 insulated rigid tank contains 1.54 kg of carbon dioxide at 100 kPa. Now paddle-wheel work is done on the system until the pressure in the tank rises to 135 kPa. Determine (a) the actual paddle-wheel work done during this process and (b) the minimum paddle-wheel work with which this process (between the same end states) could be accom- plished. Take T0

W = 100.44 KJ

W(minimum) = 7.23KJ

Explanation:

Using Ideal gas

Given that ,

P₁ and P₂ = (100 and 135) Kpa respectively

V = 0.8m³

m = 1.54kg

R = 0.1889

T₁ =

   =

   = 275k

T₂ =

    =

    =  371k

The energies at initial and final state are determined for the given temperature using interpolation and were divided by molar mass of carbon dioxide

Given that

μ = internal energies

μ₁ = 141.56kjkg⁻¹

μ₂ = 206.78kjkg⁻¹

The actual work is the difference of the internal energies

W = m(μ₂ - μ₁)

   = 1.54(206.78 - 141.56)kj

   = 100.4kj

The zero entropies at the initial and final states are determined

∫ represent entropy

∫₁ = 4.788 kjkg⁻¹k⁻¹

∫₂ = 5.0478 kjkg⁻¹k⁻¹

(b)

The minimum work required is

w(minimum) = m(μ₂ - μ₁ - T₀(∫₂ - ∫₁))

                    = w - mT₀ (∫₂ - ∫₁ - RIn

                    = 100kj - 1.54 * 298 (5.0478 - 4.788 - 0.1889 * In) kj

                    = 7.23 kj

d.7.5cm is your answer

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