One mole of a monatomic ideal gas undergoes a transformation from an initial state with temperature 290 K and volume 30 litres to a final state with temperature 310 K and volume 16 litres. On the pressure–volume ( P - V ) diagram, this process is represented by a straight line path. The magnitude of the work done (in joules) during this process is close to

Step-by-step solution

For a straight line path on the P-V diagram, the work done is given by the area of the trapezium under the curve: W = 1 2 (P₁ + P₂)(V₂ - V₁) The magnitude of the work done is: |W| = 1 2 (P₁ + P₂)|V₂ - V₁| Using the ideal gas equation P = nRT V for n = 1 mole: P₁ = RT₁ V₁ = 8.314 290 30 10⁻³ = 80.368 10^3 Pa P₂ = RT₂ V₂ = 8.314 310 16 10⁻³ = 161.083 10^3 Pa The change in volume is: |V₂ - V₁| = |16 - 30| 10⁻³ = 14 10⁻³ m ^3 Substituting these values into the work done formula: |W| = 1 2 (80.368 10^3 + 161.083 10^3) 1