Answer :

For an isothermal process, the ideal gas equation is given as


PV = constant … (i),


Where


P = pressure


V = volume.


Differentiating on both sides of (i), we get


PdV + VdP = 0


On solving for , we get


… (ii)


For a graph of P versus V, dP/dV indicates the slope.


Hence, for an isothermal process, the slope of the p-V diagram is given by -P/V.


Now for an adiabatic process, the ideal gas equation is


PVγ= constant … (iii),


where


P = pressure,


V = volume,


γ = ratio of specific heat capacities at constant pressure and constant volume.


Differentiating both sides of (ii), we get


V𝛾dP + 𝛾V𝛾-1PdV = 0 which gives


… (iv)


Hence, for an adiabatic process, the slope of the p-V diagram is given by -𝛾P/V.


Since 𝛾 > 1, we find that 𝛾P/V is greater than P/V, which concludes that slope of p-V diagram of an adiabatic process is steeper than that of an isothermal process(proved).


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