# Reversible adiabatic expansion -- calculate change in E

1. Jan 8, 2017

### binbagsss

1. The problem statement, all variables and given/known data

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

2. Relevant equations

$pV^{\gamma} = constant$*, where $\gamma=\frac{c_p}{c_v}$

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

3. 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)

2. Jan 8, 2017

### vela

Staff Emeritus
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?

3. Jan 8, 2017

### 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.

4. Jan 8, 2017

### binbagsss

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

What I dont understand is what I asked.

5. Jan 8, 2017

### vela

Staff Emeritus

6. Jan 8, 2017

### binbagsss

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.

7. Jan 8, 2017

### vela

Staff Emeritus

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?

8. Jan 11, 2017

### binbagsss

temperature

9. Jan 11, 2017

### TSny

(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 $\gamma$.