A gas is confined within a cylindrical vessel equipped with a frictionless piston. The gas is heated slowly for some time. Throughout the process, 10 J of heat is added, and the piston moves outward by 10 cm . Determine the increase in the internal energy of the gas. The cylinder's cross-sectional area is 4 cm ^2 , and the atmospheric pressure is 100 kPa .

Step-by-step solution

The process is isobaric since the piston is frictionless and moves slowly, maintaining equilibrium with the atmospheric pressure. The work done by the gas is given by W = P V = P A d . Substituting the given values: W = (100 10^3 Pa ) (4 10⁻⁴ m ^2) (0.1 m ) = 4 J According to the first law of thermodynamics: Q = U + W Given that the heat added is Q = 10 J , the increase in internal energy U is: U = Q - W = 10 J - 4 J = 6 J