Q.For a given ideal gas, $6\times 10^5\ \text{J}$ of heat energy is supplied and the volume of the gas is increased from $4\ \text{m}^3$ to $6\ \text{m}^3$ at atmospheric pressure. Calculate

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. …