contestada

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,

Respuesta :

batolisis

Answer:

The Internal energy of the gas did not change

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

Otras preguntas