The work done on an ideal gas system in an isothermal process is -400J. What is the change in internal (thermal) energy of the gas? (a) 0J (b) -400J (c) 400J (d) 200 J
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Answered on
Given :
Work done = -400 J
ΔU = Q + W
Here,
ΔU = Total internal energy
Q = heat
W = work done
Since, it is an isothermal process, the temperature is constant. So, there is no change in heat. Which also means that there is no change in internal energy of the gas.
Hence, option (A) is correct.