Q. 7

# Show that the slo

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

PV^{γ}= 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

V^{𝛾}dP + 𝛾V^{𝛾}^{-1}PdV = 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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