A gas undergoes a process where its pressure varies linearly with volume from 10 kPa , 200 cc to 50 kPa , 50 cc . Assuming no heat is added to or extracted from the gas, find the change in the internal energy of the gas.

Step-by-step solution

Given initial state: P₁ = 10 kPa = 10 10^3 Pa V₁ = 200 cc = 200 10⁻⁶ m ^3 Final state: P₂ = 50 kPa = 50 10^3 Pa V₂ = 50 cc = 50 10⁻⁶ m ^3 Since the pressure varies linearly with volume, the process on the P-V diagram is a straight line. The work done by the gas is the area under the P-V graph, which forms a trapezium: W = 1 2 (P₁ + P₂)(V₂ - V₁) Substituting the values: 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) = -4.5 J According to the first law of thermody