Question

Heat flows into a gas in a piston and work is performed on the gas by its surroundings. The amount of work done is equal to the heat added. In this situation,

Answer 1

The Internal energy of the gas did not change

Step-by-step explanation:

In this situation the Internal energy of the gas did not change and this is because according the the first law of thermodynamics

Δ U = Q - W ------ ( 1 )

Δ U = change in internal energy

Q = heat added

W = work done

since Q = W. the value of ΔU will be = zero i.e. No change