A gas undergoes a process where its pressure varies linearly with volume from 10 kPa , 200 cc to 50 kPa , 50 cc . Determine the work done by the gas.

Step-by-step solution

The work done by the gas in a process where pressure varies linearly with volume is given by the area under the P-V graph, which forms a trapezium. Initial pressure P₁ = 10 kPa = 10 10^3 Pa Final pressure P₂ = 50 kPa = 50 10^3 Pa Initial volume V₁ = 200 cc = 200 10⁻⁶ m ^3 Final volume V₂ = 50 cc = 50 10⁻⁶ m ^3 The work done W is calculated as: W = 1 2 (P₁ + P₂)(V₂ - V₁) W = 1 2 (10 10^3 + 50 10^3)(50 10⁻⁶ - 200 10⁻⁶) W = 1 2 (60 10^3)(-150 10⁻⁶) W = 30 10^3 (-150 10⁻⁶) W = -4500 10⁻³ J W = -4.5 J