Question 41. A thermodynamic system is taken from an original state A to an intermediate state B by a linear process as shown in the figure. Its 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 is :
(1) 33800 J
(2) 2200 J
(3) 800 J
(4) 1200 J
Answer: 800 J
Explanation:
Work done by the gas is equal to the area under the P–V graph.
From A to B, the process is linear. So, work done from A to B is the area of a trapezium.
Average pressure between A and B is \frac{8000+6000}{2} dyne/cm²
Change in volume from A to B is 4 m³
So, work done from A to B is
W_{AB}=\frac{1}{2}(8000+6000)\times4From B to C, the process is isobaric. Pressure at B is 4000 dyne/cm²
Volume decreases by 4 m³
So, work done from B to C is W_{BC}=-4000\times4
Total work done is W=W_{AB}+W_{BC}
W=\left[\frac{1}{2}(8000+6000)-4000\right]\times4W=2000\times4 dyne·cm
Convert dyne/cm² to N/m², 1 dyne/cm² =10^{-1} N/m²
So, W=2\times10^{3}\times10^{-1}\times4
W=8\times10^{2} J
W=800 J