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.
You need to login to perform this action.
You will be redirected in
3 sec