Answer:
Statements A & B are true.
Explanation:
Heat is converted completely into work during isothermal expansion.
Work done in an isothermal process is given as:
[tex]W=n.R.T.ln(\frac{V_f}{V_i} )[/tex]
where the subscripts denote final and initial conditions.
In case of an ideal gas the change in internal energy is proportional to the change in temperature. Here the temperature is constant in the process, therefore ΔU=0.
From the first law of thermodynamics:
[tex]dW=dQ+dU[/tex]
for isothermal process it becomes
[tex]dW=dQ[/tex]
Isothermal expansion is reversible under ideal conditions.
Ideally there is no friction involved as all the heat is converted into work, so the process is reversible.
During the process of isothermal expansion, the gas does more work than during an isobaric expansion (at constant pressure) between the same initial and final volumes.
As we know that the work done is given by the area under the P-V curve so in case of the work done between two specific states of volume by an isothermal process we have only one path and by isobaric process we have 2 paths among which one is having the higer work done than the isothermal process nd the other one is having the lesser area under the curve than that of under isothermal curve.