Reversible adiabatic expansion -- calculate change in E

[binbagsss]

Homework Statement

Calculate the change in internal energy of an ideal gas the expands reverisbly and adiabatically from $v_i$ to $v_f$

Homework Equations

$pV^{\gamma} = constant$*, where $\gamma=\frac{c_p}{c_v}$
for a reversible adiabatic.

$E=\alpha p V$, $\alpha$ a constant.

The Attempt at a Solution

The solution does

$E=\alpha pV$

$\implies$ $E_f-E_i=\alpha (p_fV_f-p_iV_i) = W$

(And then continues to eliminate a variable using * which I am fine with etc)

MY QUESTION IS

(by the first law) $dE=dW=-pdV$

But I don't understand how we can simply do $p_fV_f-p_iV_i$ because I thought that, in general, $p=p(V)$ and so we need to integrate over $p$ instead of just looking at the initial and final state.(unless a process is isochoric - $p(vf-vi)=W$).

Or in general would this be the case but due to * we do not need to integrate ? (if so how can this be seen)

Many thanks in advance.

 

[vela]

You have $p_iV_i^\gamma = k = p_fV_f^\gamma$, so the work done is
$$W = -k\int_{V_i}^{V_f} V^{-\gamma}\,dV = -k\frac{V_f^{-\gamma+1}}{-\gamma+1}+k\frac{V_i^{-\gamma+1}}{-\gamma+1}.$$ Replace the first instance of $k$ with $p_fV_f^\gamma$ and the second with $p_iV_i^\gamma$. What do you end up with?

 

[TSny]

You can find W by integrating PdV. Following Vela, you should find that you get the correct result.

However, the first law tells you that for this process W must equal ΔE. Finding ΔE does not require any integration.

 

[binbagsss]

You have $p_iV_i^\gamma = k = p_fV_f^\gamma$, so the work done is
$$W = -k\int_{V_i}^{V_f} V^{-\gamma}\,dV = -k\frac{V_f^{-\gamma+1}}{-\gamma+1}+k\frac{V_i^{-\gamma+1}}{-\gamma+1}.$$ Replace the first instance of $k$ with $p_fV_f^\gamma$ and the second with $p_iV_i^\gamma$. What do you end up with?

I have seen this method of doing it, and I understand that, that is fine.

What I don't understand is what I asked.

 

[vela]

If you follow through with my suggestion, you end up with the result you're asking about. How does that not answer your question?

 

[binbagsss]

If you follow through with my suggestion, you end up with the result you're asking about. How does that not answer your question?

that tells me that you are doing is correct.the two methods are the same

it doesn't tell me how or why it is okay to not need a $p=p(v)$ and integrate.

 

[vela]

it doesn't tell me how or why it is okay to not need a $p=p(v)$ and integrate.

If you're going to calculate the work to find $\Delta E$, then you do need to do exactly that.

As TSny noted, the other way is to find $\Delta E = E_f - E_i$ directly. What does the internal energy of an ideal gas depend on?

 

[binbagsss]

If you're going to calculate the work to find $\Delta E$, then you do need to do exactly that.

As TSny noted, the other way is to find $\Delta E = E_f - E_i$ directly. What does the internal energy of an ideal gas depend on?

temperature

 

[TSny]

But I don't understand how we can simply do $p_fV_f-p_iV_i$ because I thought that, in general, $p=p(V)$ and so we need to integrate over $p$ instead of just looking at the initial and final state.(unless a process is isochoric - $p(vf-vi)=W$).

(1) Integrating by parts gives $\int_i^fpdV = (pV) |_i^f - \int_i^fVdp$.

(2) For an adiabatic process, show that $Vdp = -\gamma pdV$.

Combine (1) and (2) to show that $\int_i^fpdV$ can be expressed in terms of just $(pV)|_i^f$ and $\gamma$.