Question

Prove that the slope of P-V graph for an adiabatic process is Y times that of the isothermal process.

Answer 1

Answer:

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Answer 2

The slope of the P-V graph of the adiabatic process is Y times that of the isothermal process.

• For adiabatic process

P $V^{Y}$ = constant

On differentiating both sides of the above equation.

P Y $V^{Y-1}$ dV + $V^{Y}$ dP =0

⇒ $V^{Y}$ dP = - P Y $V^{Y-1}$ dV

⇒ (dP / dV) = - Y P $V^{Y-1}$/ $V^{Y}$

The slope of adiabatic process is (dP/ dV).

So, the slope of the adiabatic graph is (dP/ dV) = -Y P/ V

• For isothermal process

P V = constant

On differentiating both the sides of the above equation.

P dV + V dP = 0

⇒ V dP = - P dV

The slopr for isothermal process is (dP/ dV).

The slope of isthermal graph is (dP / dV) = -P/V

Now on calculating ratio of both slopes,

The ratio of the slope of the adiabatic graph to isothermal graph =

⇒ $\frac{-Y dP/ dV}{dP/dV}$ = -Y

Hence proved that the slope of the PV graph for an adiabatic process is Y times that of the isothermal process.