Figure shows the variation of the internal energy U with the density ρ of an ideal monoatomic gas for a thermodynamic process AB . Process AB is a part of a rectangular hyperbola. Find the work done (in joule) by gas in process AB .

Step-by-step solution

Suppose final internal energy is x Process AB is isobaric process U ρ = constant 2 × x = 5   × 2 x = 5   J In process AB Δ U = n f R Δ T 2 n = number of mole f = degree of freedom Δ T = Change in temperature 5 - 2 = n   × 3   × R   × Δ T 2 6 3 = n R Δ T 2 = n R Δ T Work done =   ∫ P d V = P Δ V = n R Δ T = 2   J