If 100 ~L of an ideal gas at a pressure of 2 ~atm compressed isothermally and reversibly to a final volume of ' X ' L releases -460.6 ~L ~atm heat, the final volume ' X ' (in L ) is

Step-by-step solution

U = q + w =0 w =- q We have, q =-460.6 ~L ~atm w =+460.6 ~L ~atm Now, aligned & W =-2.303 nRT ( V ₂ ~V ₁ ) & =-2.303 ( P ₁ ~V ₁ ) ( V ₂ ~V ₁ ) & =-2.303(2 100) ( V ₂ 100 ) & =-460.6 [ ( V ₂ )- (100) ] & -460.6 [ ( V ₂ )-2 ]=921.2-460.6 ( V ₂ )=460.6 & ( V ₂ )= 921.2-460.6 460.6 =1 & V ₂=10 ~L ( X ) aligned