NEET-XII-Physics
26: Laws of Thermodynamics
- #7Figure shows three paths through which a gas can be taken from the state A to the state B. Calculate the work done by the gas in each of the three paths.
FigureAns : Work done during any process, W = `` P∆V``
If both pressure and volume are changing during a process, then work done can be found out by finding the area under the PV diagram.
In path ACB, for line AC:
Since initial volume is equal to final volume, `` ∆V=0``.
`` \Rightarrow ``WAC = `` P∆V`` = 0
For line BC:
P = 30`` \times {10}^{3}`` Pa
`` {W}_{ACB}`` = WAC + WBC = 0 + P∆V
= 30 × 103 × (25 - 10) × 10-6
= 0.45 J
For path AB:
Since both pressure and volume are changing, we use the mean pressure to find the work done.
Mean pressure, P = `` \frac{1}{2}\times (30+10)\times {10}^{3}`` = 20 kPa
WAB = `` \frac{1}{2}`` × (10 + 30) × 103 × 15 × 10-6
= `` \frac{1}{2}`` × 40 × 15 × 10-3 = 0.30 J
Initial volume in path ADB, along line DB is the same as final volume. Thus, work done along this line is zero.
Along line AD, P = 10 kPa
W = WAD + WDB
= 10 × 103 (25 - 10) × 10-6 + 0
= 10 × 15 × 10-3 = 0.15 J
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