10 mole of an ideal gas is undergoing the process shown in the figure. The heat involved in the process from P₁ to P₂ is Joule ( P₁=21.7 ~Pa and .P₂=30 ~Pa , C _ v =21 ~J / K . mol , R=8.3 ~J / mol . K ) . The value of is _ _ _ _ .

Step-by-step solution

From the given P-V diagram, the process from P₁ to P₂ is an isochoric process because the volume remains constant at V = 1 m ^3 . For an isochoric process, the work done W = 0 . According to the first law of thermodynamics, the heat involved is Q = U + W = U . The change in internal energy is given by U = n C_v T . Using the ideal gas equation PV = nRT , we have n T = V P R . Substituting this into the expression for Q : Q = C_v ( V(P₂ - P₁) R ) Given values: n = 10 moles, V = 1 m ^3 , P₁ = 21.7 Pa , P₂ = 30 Pa , C