Q.One mole of an ideal gas is initially kept in a cylinder with a movable frictionless and massless piston at pressure of 1.01 MPa, and temperature 27 °C. It is then expanded till its volume is doubled. How much work is done if the expansion is isobaric? (R = 8.314 SI Units)

An isobaric process occurs at constant pressure; temperature, volume and internal energy are all free to change. From $PV=\mu RT$ with $P$ fixed, $V\propto T$, so a state going from $(V_i,T_i)$ to $(V_f,T_f)$ satisfies $\dfrac{T_f}{V_f}=\dfrac{T_i}{V_i}$, and the V-T graph is a straight line through the origin; on a P-V diagram, an isobaric process is a horizontal line. The work done by the gas is $W = P(V_f-V_i) = P\Delta V,$ positive for expansion and negative for compression, equal in either case to the (rectangular) area under the horizontal P-V line. Using the ideal gas equation this can also be written $W=\mu RT_f\left(1-\dfrac{T_i}{T_f}\right)$. Applying the first law, $\Delta U = Q - P\Delta V$. Common real-life isobaric processes include cooking food in an open vessel (always at atmospheric pressure above the food) and a gas heated under a piston loaded with fixed weights. Comparing all four basic processes for the same volume change, the work done is greatest for an isobaric process and least (in fact, purely at the expense of internal energy) for an adiabatic process. …