Question

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 :

• A. 33800 J
• B. 2200 J
• C. 0 J
• D. 1200 J

Answer 1

Answer: (C)

0 J

Answer 2

Step 1: Calculate Work Done from A to B: - Since the process from A to B is linear on the pressure-volume diagram, the work done W AB can be calculated as the area under the line AB. - The average pressure from A to B is $\frac{8000 + 4000}{2} = 6000 \, \text{dyne/cm}^2$. - The volume change from A to B is 4 m3.

$W_{AB} = \text{Average Pressure} \times \text{Change in Volume}$

$W_{AB} = 6000 \times 4 \, \text{dyne/cm}^2 \times \text{m}^3$

Step 2: Convert Units: - Convert dyne/cm2 to N/m2 by using 1 dyne/cm2 = 10-5 N/m2.

$W_{AB} = 6000 \times 10^{-5} \times 4 \, \text{J} = 800 \, \text{J}$

Step 3: Calculate Work Done from B to C: - From B to C, the process is isobaric (constant pressure), so work done W BC = Pressure × Change in Volume. - The pressure at B and C is 4000 dyne/cm2. - Volume change from B to C is −4 m3 (since the volume is reducing).

$W_{BC} = 4000 \times (-4) \times 10^{-5} \, \text{J} = -800 \, \text{J}$

Step 4: Total Work Done:

$W_{total} = W_{AB} + W_{BC} = 800 - 800 = 0 \, \text{J}$

So, the correct answer is: 0J.