Consider an adiabatic process and an isothermal process that pass through a common point p0 , V0 in a graph of p versus V (see Fig. 18-3). Show that the slope of the adiabatic curve is greater than that of the isothermal curve at this point.
The correct answer and explanation is :
To prove that the slope of the adiabatic curve is greater than that of the isothermal curve at a common point ( p_0 ), ( V_0 ), we can begin by considering the equations for both the adiabatic and isothermal processes.
1. Equation for the Isothermal Process
For an isothermal process, the temperature remains constant. The relationship between pressure and volume for an ideal gas undergoing an isothermal process is given by:
[
pV = nRT
]
where:
- ( p ) is the pressure,
- ( V ) is the volume,
- ( n ) is the number of moles of gas,
- ( R ) is the gas constant,
- ( T ) is the constant temperature.
The slope of the isothermal curve at any point can be obtained by differentiating this equation implicitly with respect to ( V ):
[
\frac{dp}{dV} = -\frac{nRT}{V^2}
]
This gives the slope of the isothermal curve at any point ( (V_0, p_0) ) where ( p_0 = \frac{nRT}{V_0} ).
2. Equation for the Adiabatic Process
For an adiabatic process, there is no heat exchange (( Q = 0 )). The relationship between pressure and volume for an ideal gas undergoing an adiabatic process is given by:
[
pV^\gamma = \text{constant}
]
where ( \gamma ) is the adiabatic index (( \gamma = \frac{C_P}{C_V} ), with ( C_P ) being the heat capacity at constant pressure and ( C_V ) being the heat capacity at constant volume).
Differentiating this equation with respect to ( V ):
[
pV^\gamma = \text{constant} \quad \Rightarrow \quad \frac{dp}{dV} = -\gamma \frac{p}{V}
]
At the common point ( (V_0, p_0) ), the slope of the adiabatic curve is:
[
\frac{dp}{dV} = -\gamma \frac{p_0}{V_0}
]
3. Comparison of Slopes
At the point ( (V_0, p_0) ), we compare the slopes of the isothermal and adiabatic curves. The slope for the isothermal process is:
[
\frac{dp}{dV} \bigg|_{\text{isothermal}} = -\frac{nRT}{V_0^2}
]
The slope for the adiabatic process is:
[
\frac{dp}{dV} \bigg|_{\text{adiabatic}} = -\gamma \frac{p_0}{V_0}
]
We can express ( p_0 ) for the isothermal process as ( p_0 = \frac{nRT}{V_0} ). Substituting this into the expression for the adiabatic slope:
[
\frac{dp}{dV} \bigg|_{\text{adiabatic}} = -\gamma \frac{nRT}{V_0^2}
]
Now comparing the magnitudes of these slopes:
- The slope of the isothermal curve: ( \left|\frac{dp}{dV} \bigg|_{\text{isothermal}}\right| = \frac{nRT}{V_0^2} )
- The slope of the adiabatic curve: ( \left|\frac{dp}{dV} \bigg|_{\text{adiabatic}}\right| = \gamma \frac{nRT}{V_0^2} )
Since ( \gamma > 1 ) for most gases, it follows that the magnitude of the slope of the adiabatic curve is greater than that of the isothermal curve. Therefore, the slope of the adiabatic curve is steeper than that of the isothermal curve at the common point ( p_0 ), ( V_0 ).
Conclusion
At the common point ( p_0 ), ( V_0 ), the slope of the adiabatic curve is greater than that of the isothermal curve. This is because ( \gamma > 1 ), which causes the adiabatic curve to have a steeper slope at that point compared to the isothermal curve.