A thermodynamic system is taken from an original state A to an intermediate state B by a linear process as shown in the figure. It's volume is then reduced to the original value from B to C by an isobaric process. The total work done by the gas from A to B and B to C would be:
The graph plots Pressure (Dyne/cm^2) on the y-axis (with coordinates at 4000 and 8000) versus Volume (m^3) on the x-axis (with marks at 3 and 7). State path goes linearly from A to B, then horizontally from B to C.
The graph plots Pressure (Dyne/cm^2) on the y-axis (with coordinates at 4000 and 8000) versus Volume (m^3) on the x-axis (with marks at 3 and 7). State path goes linearly from A to B, then horizontally from B to C.
A.33800 ~J$33800 \mathrm{~J}$
B.2200 ~J$2200 \mathrm{~J}$
C.600 ~J$600 \mathrm{~J}$
D.1200 ~J$1200 \mathrm{~J}$
Solution & Explanation
Related Formula
Work done (W$W$) in a thermodynamic process corresponds to the area under the curve on a P-V$P-V$ diagram:
W = ∫ P dV$$W = \int P \, dV$$
Core Logic
Let's compute the work done along the distinct steps AB$AB$ and BC$BC$ matching the values shown in the graph:
The graph plots Pressure (Dyne/cm^2) on the y-axis (with coordinates at 4000 and 8000) versus Volume (m^3) on the x-axis (with marks at 3 and 7). State path goes linearly from A to B, then horizontally from B to C.
Step 1: Calculate Work Done in Process AB
Process AB$AB$ is a linear expansion. Fulfilling the area of the trapezoid underneath path AB$AB$:
Since 800 ~J$800 \mathrm{~J}$ is not present in the options, this question was declared a BONUS.
Pattern Recognition
Always be highly vigilant of mixed unit systems in thermodynamic graphics. Here, pressure is in CGS (Dyne/cm²$\text{Dyne/cm}^2$) while volume is in MKS (m³$\text{m}^3$). Converting early keeps you clear of algebraic traps.
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
