# Thermodynamics work done question

1. Oct 18, 2006

### junglep

hey guys got this question that i have been stuck on for a while.

air is expanded from 1M Pa at 327 degrees celcius to 200kPa in a closed piston cylinder device. for the process PV^1.2 = constant. calculate the work done in kJ/kg during this process

i hav managed to work out the temperature after expansion using T2/T1 = (P1/P2)^(n-1/n) but i dont know how to work out the work done without knowing the mass or any of the volumes

if work = (p1v1 - p2v2)/ 1-n

then surely i need the volumes to work out the work done

any help will be welcomed

cheers

Last edited: Oct 18, 2006
2. Oct 18, 2006

### quasar987

If the gaz can be considered ideal, then by conservation of the number of moles of gaz,

$$\nu_i = \nu_f[/itex] you must have [tex]V_f=\frac{p_fT_i}{p_iT_f}V_i$$

So

$$W=\int_{V_i}^{\frac{p_fT_i}{p_iT_f}V_i}pdV = \int_{V_i}^{\frac{p_fT_i}{p_iT_f}V_i} \frac{\alpha}{V^{1.2}}dV$$

And substitude back $\alpha = p_iV_i^{1.2}$ at the end.

3. Oct 18, 2006

### junglep

but i dont know any of the volumes so this method would not work

4. Oct 18, 2006

### Andrew Mason

If $PV^\alpha = K$ where $\alpha = 1.2$ (note: this is not the $\gamma$ for air which is 1.4), then substituting V = nRT/P gives:

$$P^{1-\alpha}T^\alpha = K/n^\alpha R^\alpha = K'$$

So:

$$P_1^{1-\alpha}T_1^\alpha = P_2^{1-\alpha}T_2^\alpha = K'$$

From that, work out PdV in terms of K' and T and integrate from T1 to T2

AM

Last edited: Oct 18, 2006
5. Oct 18, 2006

### quasar987

6. Oct 19, 2006

### Andrew Mason

Find T2 from the relationship:

$$P_1^{(1-\alpha)}T_1^\alpha = P_2^{(1-\alpha)}T_2^\alpha$$

so:

$$T_2 = \left(P_1^{(1-\alpha)}T_1^\alpha/P_2^{(1-\alpha)}\right)^{1/\alpha}$$

Use PV=nRT to find V:

$$V_1 = nRT_1/P_1$$

$$V_2 = nRT_2/P_2$$

Integrating PdV from V1 to V2 using $P = K/V^\alpha$:

$$W = \int_{V_1}^{V_2} PdV = \int_{V_1}^{V_2} KdV/V^\alpha$$

You just have to work that out.

AM

Last edited: Oct 19, 2006
7. Oct 19, 2006

### junglep

what is n in the equation

pv = nRT?

i thought the perfect gas eqn was pv = mass * R * T

also i am not given a value for the gas constant (R). if it is any help the answer that is given in the book is in kJ/kg not in J.

8. Oct 19, 2006

### Andrew Mason

n is the number of moles of the gas. R is in units of J/mole K.

This problem does not give you n or V, so assume n = 1 in which case: PV = MRT where M is the mass of one mole of air (29 g/mole). Essentially, you are working out and using the volume for one mole of air.

AM