Which curve is more steeper, isothermal or adiabatic? Explain please

1 Answer
Morgan
Mar 25, 2018

Adiabats are steeper than isotherms.

Explanation:

When we graph adiabatics and isotherms, we might do so on a P vs. V diagram, or a pressure vs. volume diagram. Therefore, the slope of any curve on the graph is #Deltay//Deltax=DeltaP//DeltaV#.

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For an isothermal process, we know that there is no change in temperature, i.e. #DeltaT=0#. Then, by the ideal gas law, we have:

#PV="constant"#

Let's say #PV=c#, and separate our variables:

#=>P=c/V#

Now we'll take the derivative to look at the change in the variables (derivative = slope):

#=>(dP)/(dV)=-c/V^2#

Let's rewrite our equation:

#=>(dP)/(dV)=-c/V*1/V#

From above #P=c/V#, so making this substitution, we have:

#=>(dP)/(dV)=-P/V#

For an adiabatic process, the energy input into the system by heating is necessarily zero, i.e. #Q=0#. We begin with the relationship #PV^(gamma)="constant"#

#=>PV^(gamma)=c#

Again, we solve for #P#:

#P=c/V^(gamma)#

As with the isothermal process:

#=>(dP)/(dV)=-gamma*c/(V^(gamma+1))#

Where we have added one to #gamma# because it is #V^(-gamma)#.

#=>(dP)/(dV)=-gamma*c/V^(gamma)*1/V#

#=>(dP)/(dV)=-gamma*P/V#

As gamma is always greater than 1, we see that the slope of an adiabat is greater than that of an isotherm by a factor of #gamma#.

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