CBSE-XI-Physics

27: Specific Heat Capacities of Gases

with Solutions -
  • #7
    Show that the slope of the p-V diagram is greater for an adiabatic process compared to an isothermal process.
    Ans : In an isothermal process,
    PV = k ...(i)
    On differentiating it w.r.t V, we get
    `` V\frac{\,\mathrm{\,d\,}P}{\,\mathrm{\,d\,}V}+P=0``
    `` \frac{\,\mathrm{\,d\,}P}{\,\mathrm{\,d\,}V}=-\frac{P}{V}``
    `` \frac{\,\mathrm{\,d\,}P}{\,\mathrm{\,d\,}V}=-\frac{k}{{V}^{2}}[\text{U}\,\mathrm{\,sing\,}(\,\mathrm{\,i\,}\left)\right],\text{k = constant}``
    k = constant
    In an adiabatic process,
    `` P{V}^{\gamma }=K....\left(\,\mathrm{\,ii\,}\right)``
    On differentiating it w.r.t V, we get
    `` {V}^{\gamma }\frac{\,\mathrm{\,d\,}P}{\,\mathrm{\,d\,}V}+\gamma P{V}^{\gamma -1}=0``
    `` \frac{\,\mathrm{\,d\,}P}{\,\mathrm{\,d\,}V}=-\frac{\gamma P{V}^{\gamma -1}}{{V}^{\gamma }}``
    `` \frac{\,\mathrm{\,d\,}P}{\,\mathrm{\,d\,}V}=-\frac{\gamma K}{{V}^{\gamma +1}}[\text{U}\,\mathrm{\,sing\,}(\,\mathrm{\,ii\,}\left)\right],\gamma >1\text{and}``
    K is constant
    `` \gamma \text{and}\frac{\,\mathrm{\,d\,}P}{\,\mathrm{\,d\,}V}`` are the slope of the curve and the ratio of heat capacities at constant pressure and volume, respectively; P is pressure and V is volume of the system.
    By comparing the two slopes and keeping in mind that `` \gamma >1``, we can see that the slope of the P-V diagram is greater for an adiabatic process than an isothermal process.
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