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NEET-XII-Physics

26: Laws of Thermodynamics

with Solutions - page 5
Qstn# iv-16 Prvs-QstnNext-Qstn
  • #16
    A gas is initially at a pressure of 100 kPa and its volume is 2.0 m3. Its pressure is kept constant and the volume is changed from 2.0 m3 to 2.5 m3. Its Volume is now kept constant and the pressure is increased from 100 kPa to 200 kPa. The gas is brought back to its initial state, the pressure varying linearly with its volume. (a) Whether 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 : (a) 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-a
    Whether the heat is supplied to or extracted from the gas in the complete cycle? (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:
  • #16-b
    How 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: