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

1 Answer
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.