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 :
- 33800 J
- 2200 J
- 0 J
- 1200 J
The Correct Option is C
Approach Solution - 1
To find the total work done by the gas from state A to B and then from B to C, we need to analyze the given thermodynamic process using the principles of work done in a pressure-volume (P-V) diagram.
The work done by a gas during a process is given by the area under the curve in the P-V diagram. Let's analyze each path separately:
- Path A to B:
- This is a linear process, which means it's neither isothermal, isobaric, nor isochoric.
- To compute the work done from A to B, we need the formula for the area under the straight line in a P-V diagram, which can generally be calculated using the trapezoidal or triangular area methods.
- Path B to C:
- This process is isobaric (constant pressure).
- The work done in an isobaric process is given by the formula: \(W = P \Delta V\), where \( P \) is the constant pressure and \( \Delta V \) is the change in volume.
- From the diagram, since \( V_{\text{C}} = V_{\text{A}} \), there is no change in volume in the net process back to the original state, making \(\Delta V = 0\).
Since the volume returns to its original value, any work done from A to B is canceled out by the work done from B to C (assuming ideal conditions with no other energy loss or gain). Therefore, the total work done over the complete cycle A to B to C is:
\(W_{\text{total}} = W_{\text{AB}} + W_{\text{BC}} = 0\)
Hence, the correct answer is 0 J.
Approach Solution -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 WAB 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 WBC = 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.
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