The internal energy of a monoatomic gas is 3 nRT. One mole of helium is kept in a cylinder having internal cross section area of 17 ~cm ² and fitted with a light movable frictionless piston. The gas is heated slowly by suppling 126 J heat. If the temperature rises by 4^ C , then the piston will move _ _ _ _ cm. (atmospheric pressure =10⁵ ~Pa )

Step-by-step solution

The question states the internal energy of the gas is U = 3nRT . The change in internal energy is U = 3nR T . Given: n = 1 mole, T = 4^ C = 4 K, and R 8.314 J/mol·K. U = 3 1 8.314 4 = 99.768 J. According to the first law of thermodynamics, Q = U + W . Given heat supplied Q = 126 J. Work done by the gas W = Q - U = 126 - 99.768 = 26.232 J. Since the piston is light and movable, the process occurs at constant atmospheric pressure P = 10^5 Pa. Work done W = P V = P(A d) , where A is the cross-section area and d is the