An ideal gas undergoes a process where its pressure changes linearly with volume. The initial volume and pressure of the gas are 100 cc and 200 kPa , respectively. The gas expands to a final volume of 300 cc . If no heat is supplied to or extracted from the gas and the change in its internal energy is -60 J , the final pressure of the gas is:

Step-by-step solution

Given, Q = 0 and U = -60 J . From the first law of thermodynamics, Q = U + W 0 = -60 J + W W = 60 J The work done is the area under the P-V curve. Since the pressure changes linearly with volume, the area is a trapezium. W = 1 2 (P₁ + P₂)(V₂ - V₁) Here, V₂ - V₁ = 300 cc - 100 cc = 200 cc = 200 10⁻⁶ m ^3 = 2 10⁻⁴ m ^3 60 = 1 2 (200 10^3 + P₂ 10^3) (2 10⁻⁴) 60 = (200 + P₂) 10⁻¹ 600 = 200 + P₂ P₂ = 400 kPa Answer: 400 kPa