NEET-XII-Physics
26: Laws of Thermodynamics
- #16-aWhether the heat is supplied to or extracted from the gas in the complete cycle? (b) How much heat was supplied or extracted? (b) How much heat was supplied or extracted?Ans : Given:
P1 = 100 kPa,
V1 = 2 m3
`` {V}_{2}`` = 2.5 m3
∆V = 0.5 m3
Work done, W = P∆V
`` W=100\times {10}^{3}\times 0.5``
`` W=5\times {10}^{4}\,\mathrm{\,J\,}``
WAB = Area under line AB = 5`` \times ``104 J
If volume is kept constant for line BC, then ∆V = 0.
WBC = P∆V = 0
Work done while going from point B to C = 0
When the system comes back to the initial point A from C, work done is equal to area under line AC.
WCA = Area of triangle ABC + Area of rectangle under line AB
Total work done, W = Area enclosed by the ABCA
W = WAC `` -``WAB
From the graph, we see that the area under AC is greater than the area under AB. We also see that heat is extracted from the system as change in the internal energy is zero. (b) Amount of heat extracted = Area enclosed under ABCA
`` =\frac{1}{2}\times 0.5\times 100\times {10}^{3}=25000\,\mathrm{\,J\,}``
Page No 63: (b) Amount of heat extracted = Area enclosed under ABCA
`` =\frac{1}{2}\times 0.5\times 100\times {10}^{3}=25000\,\mathrm{\,J\,}``
Page No 63:
- #16-bHow much heat was supplied or extracted?Ans : Amount of heat extracted = Area enclosed under ABCA
`` =\frac{1}{2}\times 0.5\times 100\times {10}^{3}=25000\,\mathrm{\,J\,}``
Page No 63: