11th Class Physics Thermodynamical Processes

  • question_answer 36)
                      Consider that an ideal gas (n moles) is expanding in a process given by \[P=f(V),\] which passes through a point \[({{V}_{0}},\,{{P}_{0}})\]. Show that the gas is absorbing heat at \[({{P}_{0}},\,{{V}_{0}})\]  if the slope of the curve \[P=f(V)\] is larger than the slope of the adiabat passing through \[({{P}_{0}},\text{ }{{V}_{0}})\].                

    Answer:

                      Slope of curve, \[P=f(V)\]                 Slope of curve P at \[({{V}_{0}},\,{{P}_{0}})=f({{V}_{0}})\]             \[=-\left( \frac{{{P}_{0}}}{{{V}_{0}}} \right)\]                 Slope of adiabatic curve at \[({{V}_{0}},\,{{P}_{0}})\]             \[=-\gamma \left( \frac{{{P}_{0}}}{{{V}_{0}}} \right)\]                 According to first law of thermodynamics, heat absorbed,\[dQ=dU+pdV=n{{C}_{v}}dT+pdV\].. (i)                 Now \[PV=nRT\] or \[T=\frac{PV}{nR}\]                 \[\therefore \]\[dT=\frac{1}{nR}[PdV+VdP]\]                 Since \[P=f(V)\] and \[dP=f'(V)dV\]                 \[\therefore \] \[dT=\frac{1}{nR}[f(V)dV+Vf'(V)dV]\]                 \[=\frac{1}{nR}[f(V)+Vf'(V)dV]\]           ? (ii)                 \[\therefore \] \[dQ=n{{C}_{V}}\times \frac{1}{nR}[f(V)+V\,f'(V)]dV+PdV\]                 \[\therefore \] \[\frac{dQ}{dV}=\frac{{{C}_{V}}}{R}[f(V)+Vf'(V)]+P\]                 \[=\frac{{{C}_{v}}}{R}[f(V)+Vf'(V)]+f(V)\]                 When \[V={{V}_{0}}\]                 \[\frac{dQ}{dV}=\frac{{{C}_{\upsilon }}}{({{C}_{p}}-{{C}_{\upsilon }})}[f({{V}_{0}})+{{V}_{0}}f'({{V}_{0}})]+f({{V}_{0}})\]                 \[(\because \,{{C}_{p}}-{{C}_{\upsilon }}=R)\]                 Since \[\frac{{{C}_{p}}}{{{C}_{\upsilon }}}=r\]                 \[\therefore \] \[{{\left. \frac{dQ}{dV} \right|}_{V={{V}_{0}}}}=\left( \frac{1}{\gamma -1}+1 \right)f({{V}_{0}})+\frac{{{V}_{0}}f'({{V}_{0}})}{\gamma -1}\]                 \[=\frac{\gamma }{\gamma -1}f({{V}_{0}})+\frac{{{V}_{0}}}{\gamma -1}f'({{V}_{0}})\]             Since \[f({{V}_{0}})={{P}_{0}}\]                 \[\therefore \] \[\frac{dQ}{dV}=\frac{\gamma {{P}_{0}}}{\gamma -1}+\frac{{{V}_{0}}}{\gamma -1}f'({{V}_{0}})\]                 When heat is absorbed, \[\frac{dQ}{dV}>0\]                 \[\frac{1}{\gamma -1}[\gamma {{P}_{0}}+{{V}_{0}}f'({{V}_{0}})]>0\]                 or \[{{V}_{0}}f'({{V}_{0}})>-\gamma {{P}_{0}}\]                 or \[f'({{V}_{0}})>-\gamma \left( \frac{{{P}_{0}}}{{{V}_{0}}} \right)\] or \[f'({{V}_{0}})>\] slope of adiabatic curve.                 Thus, when gas absorbs heat, slope of curve is greater than the slope of adiabatic curve.                


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