How do you calculate the work done in an adiabatic process?

[pivoxa15]

How do you calculate the work done in an adiabatic process?

I know change in thermal energy = work done.

which amounts to (f/2)Nk(change in T). But how do you get change in T?

Or is there another way to compute the work directly. The trouble is both the volume and temperture vary during the process.

 

[Andrew Mason]

How do you calculate the work done in an adiabatic process?

I know change in thermal energy = work done.

which amounts to (f/2)Nk(change in T). But how do you get change in T?

Or is there another way to compute the work directly. The trouble is both the volume and temperture vary during the process.

(1) $$W = \int PdV = nR\int TdV/V$$ (substituting P = nRT/V)

But the adiabatic condition $PV^\gamma = nRTV^{\gamma-1} = K$ applies, so (1) becomes:

$$W = K\int dV/V^{\gamma}$$

Determine that integral for a particular volume change and that is the work.

AM

 

[pivoxa15]

But do you think it is easier to find the change in internal energy hence only change in temperture between the two states? This will give the work as Q=0.

 

[Andrew Mason]

But do you think it is easier to find the change in internal energy hence only change in temperture between the two states?

You could take that approach. You apply the adiabatic condition in terms of temperature and integrate $\int nC_vdT$ from initial to final temperature. But you have to apply the adiabatic condition to find the initial and final temperatures.

That will give you the change in internal energy. You then equate that to the work done (since, as you point out, Q = 0, $\Delta U = \Delta W$.

But because you have to use the adiabatic condition $T = K/nRV^{\gamma-1}$ it is essentially the same calculation that I have set out above.

The integral:

$$K \int_{V_i}^{V_f} V^{-\gamma}dV$$

is not that difficult to work out using antiderivative of $V^{-\gamma}$ = $V^{-\gamma +1}/(-\gamma + 1)$

$$W = K(V_f^{1-\gamma} - V_i^{1-\gamma})/1-\gamma$$

AM

 

[pivoxa15]

You could take that approach. You apply the adiabatic condition in terms of temperature and integrate $\int nC_vdT$ from initial to final temperature. But you have to apply the adiabatic condition to find the initial and final temperatures.

AM

I don't take the integral. I just use the adiabat relation to find the initial and end tempertures in terms of one temperture that I know the value to such as the initial temperture.

 

[Andrew Mason]

I don't take the integral. I just use the adiabat relation to find the initial and end tempertures in terms of one temperture that I know the value to such as the initial temperture.

The integral evaluates to $$nC_v(T_f - T_i)$$. Is that not what you use? How do you find the beginning and end temperatures?

AM

 

[Andrew Mason]

The integral evaluates to $$nC_v(T_f - T_i)$$.

Just to do the math to make my point clearer:

$$T= K/nRV^{\gamma-1}$$

So:

(1) $$nC_v(T_f - T_i) = nC_v\frac{K}{nR}(V_f^{1-\gamma} - V_i^{1-\gamma})$$

But $$C_v = C_p-R = C_v\gamma - R$$ so

$$C_v = \frac{R}{\gamma-1}$$

Substituting in (1):

$$nC_v(T_f - T_i) = \frac{K}{(\gamma-1)}(V_f^{1-\gamma} - V_i^{1-\gamma})$$

Thus:

$$\Delta U = -\Delta W$$

AM

 

[talaroue]

I have the same problem but is K a constant?

 

[talaroue]

Oh I wasn't sure! Thanks!

 

[talaroue]

is that little symbol 7/5 for diatomic? and 5/3 for monatomic?

i got a really small number...

 

[queenofbabes]

whoops. silly me. k is just a constant, from adiabatic condition $pV^\gamma$=constant

 

[talaroue]

let me try that and see how that works.

 

[talaroue]

hold on i have 2 differnt pressures and volumes and i have K=pV^alpha...which do i use or what do I do.

 

[talaroue]

https://www.physicsforums.com/showthread.php?t=323141"

 

[talaroue]

there is my problem...