Question

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

Answer 1

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`.

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`.