Heat is given to an ideal gas in an isothermal process. Then A. internal energy of the gas will decrease. B. internal energy of the gas will increase. C. internal energy of the gas will not change. D. the gas will do negative work.

Step-by-step solution

Isothermal process implications: For an ideal gas, internal energy U depends solely on temperature T , so with T = 0 , it follows that U = 0 . Applying the first law of thermodynamics, U = Q - W , yields 0 = Q - W or Q = W . Given Q > 0 (heat is added), W > 0 , indicating the gas performs positive work (expands). Internal energy remains unchanged. Final answer: B