# Calculating work in Adiabatic compressions

1. Oct 19, 2007

### T-7

Hi,

I'm thinking about how to derive an expression for the final temperature (call it T2) and the work done (call it W) in an adiabatic compression, when you only know the initial and final pressures (call them P1 and P2), and the initial temperature (call it T1). I'm not sure about my expression though, and would appreciate someone having a look at them.

Since it's an adiabatic process, I infer that

$$PV^{K} = const$$

Using the ideal gas law, and rearranging a bit, I make it that

$$T_{2} = \frac{P_{1}^{1/K-1}}{P_{2}^{1/K-1}}T_{1}$$

Is that sensible (?).

For the last bit, I ended up deriving the expression

$$W = nR\frac{T_{1}-T_{2}}{1-1/K}$$

which to me just looks too simple.

I started with the usual

$$W = \int^{V_{1}}_{V_{2}}P dV$$

replacing dV with $$-nRT\frac{dP}{P^{2}}$$ and the T with $$bP^{1-1/K}$$ since $$b^{1/K-1}T = const. = b \Rightarrow T = bP^{1-1/K}$$ to obtain, in the end, the integral:

$$W = -nRb \int^{P_{2}}_{P_{1}} P^{-1/K} dP$$

On integrating, I find

$$W = nR \frac{bP_{1}^{1-1/K}-bP_{2}^{1-1/K}}{1-1/K}$$

and then substituting for the b's in terms of T1 and P1 or T2 and P2, I found they disappeared, leaving me with

$$W = nR\frac{T_{1}-T_{2}}{1-1/K}$$

But I'm not sure that's right (?). How does it look to you?

Cheers!

2. Oct 19, 2007

### dynamicsolo

I agree with this portion.

I agree with this up to the dimensionless denominator. Since this is positive, as
$$T_{1} < T_{2}$$ and 1 < (1/K) , I take this to be your result for work done on the gas.

I chose not to change the variable of integration, so the work integral becomes

$$W = \int^{V_{2}}_{V_{1}} P_{1}(\frac{V_{1}^{K}}{V^{K}}) dV$$ .

$$W = \frac{P_{1}V_{1}^{K}}{1-K}·(V_{2}^{1-K}-V_{1}^{1-K})$$ .

Distributing the $$V_{1}^{K}$$ factor, we obtain

$$W = \frac{P_{1}}{1-K}·(V_{1}^{K}V_{2}^{1-K}-V_{1})$$ .

Returning to the adiabatic relation for a moment, we derive

$$T_{1}V_{1}^{K-1} = T_{2}V_{2}^{K-1}$$ , giving

$$V_{2}^{1-K} = \frac{T_{2}}{T_{1}}V_{1}^{1-K}$$ ,

which we can now substitute into the result from the integration to get

$$W = \frac{P_{1}}{1-K}(V_{1}^{K}\frac{T_{2}}{T_{1}}V_{1}^{1-K}-V_{1})$$ ,

or

$$W = \frac{P_{1}}{1-K}(\frac{T_{2}}{T_{1}}-1)V_{1}$$.

But $$V_{1} = \frac{nRT_{1}}{P_{1}}$$, reducing the result to

$$W = \frac{nR(T_{2}-T_{1})}{1-K}$$ .

I've calculated the work done by the gas, so we'd reverse the numerator for the work done on the gas.

Last edited: Oct 19, 2007
3. Oct 21, 2007

### T-7

I've done it your way, and get your result. And done it again my way, and still get my result! I'll try again, but surely the two approaches should end up with the same answer in the end!

4. Oct 21, 2007

### dynamicsolo

I believe a factor of 1/K may have gotten away from you. I started from

$$PV^{K} = const$$

and differentiated it to get

$$V^{K} dP + P KV^{K-1} dV = 0$$ , or

$$\frac{dP}{P} = -K \frac{dV}{V}$$ . (Logarithmic differentiation gets you here also.)

Replacing this in the integral for work done by the gas,

$$W = \int^{V_{2}}_{V_{1}}P dV$$ , gives

$$W = - (\frac{1}{K}) \int^{P_{2}}_{P_{1}} V dP$$ or

$$W = - (\frac{1}{K}) \int^{P_{2}}_{P_{1}} (\frac{c}{P^{1/K}}) dP$$ ,

where $$c = P_{1}^{1/K}V_{1}$$ .

Carrying out the integral, we obtain

$$W = -\frac{c}{K} (\frac{1}{1-\frac{1}{K}}) (P_{2}^{1-\frac{1}{K}} - P_{1}^{1-\frac{1}{K}})$$ .

Replacing c , we have

$$W = \frac{P_{1}^{1/K}V_{1}}{K-1} (P_{1}^{1-\frac{1}{K}} - P_{2}^{1-\frac{1}{K}})$$ .

If we now distribute the numerator factor through,

$$W = \frac{1}{K-1} (P_{1}V_{1} - P_{1}^{1/K}V_{1}P_{2}^{1-\frac{1}{K}})$$ .

But

$$P_{1}V_{1} = nRT_{1}$$ and, from the adiabatic relation,

$$P_{1}^{1/K}V_{1} = P_{2}^{1/K}V_{2}$$ , so the second term above is

$$P_{1}^{1/K}V_{1}P_{2}^{1-\frac{1}{K}} = P_{2}^{1/K}V_{2}P_{2}^{1-\frac{1}{K}} = P_{2}V_{2} = nRT_{2}$$ .

So, at last, the work done by the gas is

$$W = \frac{1}{K-1} (nRT_{1} - nRT_{2}) = \frac{nR}{K-1} (T_{1} - T_{2})$$ < 0 ,

the work done on the gas being the negative of this.

These manipulations get tricky (I had to check this three times to make sure I caught all the sign and distribution errors, to say nothing of fixing the TeX... >:/ ). I preferred doing this one by integrating on volume because there was less to transform.

Last edited: Oct 21, 2007
5. Oct 22, 2007

### T-7

Hi,

Ah. Now I said:

$$PV = nRT \Rightarrow dV = -\frac{nRT}{P^{2}} dP$$

If I rearranged that into the form above, it would be

$$\frac{dP}{P^{2}} = - \frac{dV}{nRT} = -\frac{dV}{PV} \Rightarrow$$

$$\frac{dP}{P} = - \frac{dV}{V}$$

And there we seem to have the discrepancy in our two approaches. You have a K in your expression, I don't. Much as I admire your proof, I don't intend to copy it. ;-) I'm not sure why I can't work out the derivative in the way that I have: the ideal gas law should hold. In my case I kept it as

$$PV = nRT \Rightarrow dV = -\frac{nRT}{P^{2}} dP$$

and rewrote the T term as a function of the pressure using the appropriate adiabatic relation. I'm not clear why that wouldn't work.

6. Oct 22, 2007

### dynamicsolo

OK, here's the problem: this differentiation would be fine for an isothermal process, but T is not constant in an adiabatic one. So you'd need to start from

V = nRT/P

and differentiate with respect to both P and T:

dV = (nR/P) dT - (nRT/[P^2]) dP .

Thermodynamics problems often make for headaches this way, because any of the quantities can be variables, so you have to keep in mind the properties of the particular process you're modeling. [Examples in basic textbooks frequently just illustrate certain constrained processes, such as isothermal or isobaric ones, so students can be misled into thinking that one or another derivative will always become zero.] It's best to start from the defining relation for the process you're analyzing, rather than the ideal gas law, which is really extremely general (and thank goodness!).

7. Oct 22, 2007

### T-7

Gotcha. 10/10 for your help on this problem. I really feel I've learned something working through this with you.

Cheers.