Question

A pressure versus volume (P-V) diagram for a system is shown in the figure. The arrows of the curve indicate the direction of the process, and the points of interest are labeled. The values for the points in the diagram are shown in the table. Pressure (Pa) Volume (m) Vo = 27.0 Vi = 20.2 V2 = 17.4 Vj = 14.2 V = 14.2 Vs = 7.51 Pressure (Pa) Po = 1.37 104 P1 = 1.37 104 P2 = 6.18 x 103 P3 = 6.18 x 10 P4 = 3.06 x 10 Ps= 1.57 x 10 Calculate the amount of work done on the system from 0-2 (W.2) and then for the entire curve from 0-5 (W05). Volume (m)

Answer 1

For Work done on the system from 0 to 2

Pressure (fa) a volume (m3)

Work done on the gas from 0 to 2 is equal to the area under the PV graph from 0 to 2

W₀₂ = W₀₁ + W₁₂

W₀₂ = area of rectangle from 0 to 1 + area of trapezium from 1 to 2

W₀₂ = 1.37×10⁴×(-6.8) + (1/2)×(6.18×10³ + 1.37×10⁴)×(-2.8)

W₀₂ = - 12.1×10⁴ J

For work done from 0 to 5

Pressure (Pa) volume (m3)

W₀₅ = W₀₁ + W₁₂ + W₂₃ + W₃₄ + W₄₅

As there is no change in volume from 3 to 4 , therefore W₃₄ = 0

W₀₅ = ( W₀₁ + W₁₂) + W₂₃ + 0 + W₄₅

Since, W₀₁ + W₁₂ = - 12.1×10⁴ J  

W₀₅ = - 12.1×10⁴ + W₂₃ + W_{​​​​​​45}

Since W₂₃ = area under the PV graph between 2 and 3

W_{​​​​45} = area under the PV graph between 4 to 5

W₀₅ = - 12.1×10⁴ + (-19.78×10³) + (- 15.49×10³)

W₀₅ = - 156.27 × 10³ J

W₀₅ = - 15.63 × 10⁴ J