Help with isotherms and adiabatic curves

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    Adiabatic Curves
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An adiabatic curve for an ideal gas is always steeper than an isotherm at the same point due to the nature of heat transfer; in an adiabatic process, no heat is exchanged, leading to a temperature increase with compression. The equations for isothermal and adiabatic processes show that the slope of the adiabatic curve is influenced by the heat capacity ratio (γ), which is greater than one. For van der Waals gases, the heat capacity ratio may not hold constant as it does for ideal gases, affecting the slope comparison. Understanding that isothermal processes maintain constant temperature while adiabatic processes do not is crucial for grasping these concepts. This distinction clarifies why the slopes differ at a given point.
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This is to help me understand a problem in the book, I don't want to state the question I'm attempting to solve though I would rather just get a nudge in the right direction conceptually. (I am analyzing an isotherm and adiabatic curve at the same point.)

My book states that for an ideal gas, an adiabatic curve at any point is always steeper than an isotherm passing through the same point: why is this?

Also, the question I'm given does not explicitly state that I'm dealing with an ideal gas, is the ratio of heat capacities still valid for a van der waals gas (which is the only other type of gas that I know how to deal with)? What about the rule i stated above concerning the slopes at a particular point?
 
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channel1 said:
My book states that for an ideal gas, an adiabatic curve at any point is always steeper than an isotherm passing through the same point: why is this?


The equation for an Isothermal Process is given by PV=\text{Constant}. Differentiate both sides to get p \,dV + v\, dP = 0 =\implies \frac{dP}{dV}= - \frac{p}{v}.

Now the equation for an Adiabatic Process is given by PV^{\gamma}=\text{Constant}. Differentiate both sides to get v^{\gamma}\, dP+\gamma \,v^{\gamma-1}\, p\, dV = 0 \implies \frac{dP}{dV}=-\gamma \frac{p}{v}.

Obviously, Since γ >1, the slope \frac{dP}{dV} for adiabatic curve is more ie., It is more steeper.
 
Ans the ratio of heat capacities won't be same for the Real gases because the Ideal Gas law itself doesn't holds.
 
A non mathematical way to look at it is to first realize that 'isothermal' means constant temperature.
If you increase the pressure then heat energy must be removed to make the change isothermal.( Increasing pressure tends to increase temperature as well as decrease volume)
In an adiabatic change no heat energy is allowed to enter or leave the system.
So in an adiabatic compression the temperature of the gas will increase which provides an additional increase in pressure.
This means that the P~V curves are steeper
 
ohhhhh ok thanks guys, together you totally gave me a complete understanding of this---so thanks a ton! :D
 

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