Why is the p.v. curve for an adiabatic process steeper than that for a

1.	Why is the p.v. curve for an adiabatic process steeper than that for an isothermal process? Explain formation of clouds at night altitude.
Answer» Explanation: For adiabatic process- PVγ=constant Differentiating above EQUATION, we have V γdP+P(γV γ−1dV)=0 ⇒V γdP=−γPV γ−1 dV ⇒ dVdP=−γ VP Hence the slope of adiabatic CURVE =−γ VP For ISOTHERMAL process- PV=constant Differentiating above equation, we have VdP+PdV=0 ⇒VdP=−PdV ⇒ dVdP=− VP Hence the slope of isothermal curve =− VP Therefore, Slope of isothermal curve Slope of adiabatic curve = (− VP )(−γVP)=γ As γ is always GREATER than 1, the slope of an adiabatic curve is greater than that of an isothermal curve by a factor of γ. Hence the adiabatic curve is STEEPER than the isothermal curve, in both the processes of expansion and compression.

Why is the p.v. curve for an adiabatic process steeper than that for an isothermal process? Explain formation of clouds at night altitude.

Answer»

Explanation:

For adiabatic process-

PVγ=constant

Differentiating above EQUATION, we have

V γdP+P(γV γ−1dV)=0

⇒V γdP=−γPV γ−1 dV

⇒ dVdP=−γ VP

Hence the slope of adiabatic CURVE =−γ VP

For ISOTHERMAL process-

PV=constant

Differentiating above equation, we have

VdP+PdV=0

⇒VdP=−PdV

⇒ dVdP=− VP

Hence the slope of isothermal curve =− VP

Therefore,

Slope of isothermal curve

Slope of adiabatic curve

= (− VP )(−γVP)=γ

As γ is always GREATER than 1, the slope of an adiabatic curve is greater than that of an isothermal curve by a factor of γ.

Hence the adiabatic curve is STEEPER than the isothermal curve, in both the processes of expansion and compression.

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