# Work done in an adiabatic process

In summary: JIn summary, the adiabatic process performed on 15 moles of an ideal gas with an initial temperature of 320K and initial volume of 0.80m3, with a final volume of 0.40m3 and an adiabatic constant of 1.44, results in a work done by the gas of -32.32kJ.

## Homework Statement

An adiabatic process is performed on 15 moles of an ideal gas. The initial temperature is 320 K and the initial volume is
0.80 m3. The final volume is 0.40 m3. The adiabatic constant for the gas is 1.44. How much work is done by the gas?

## Homework Equations

W = Integrate[p dv, vInitial, vFinal] for adiabatic processes
pV = nRT

## The Attempt at a Solution

If p(v^y) initial == p(v^y) final where y = 1.44, and v varies, then p also varies, and I don't know how to use a varying pressure with the formula for work for adiabatic processes.

I try to find the initial pressure by
pV = nRT
p*V == 15*8.31*320
With Vinitial, p = 49860Pa
With Vfinal, p = 99720Pa

And I'm stuck.

Here is my solution: for adiabatic process, you can write: PV^(gamma) = C => P = C / V^(gamma). Plug it into W = int(PdV). After that, replace C by P and V.

p*((0.8)^1.44) = c
Which gives
p = 1.37896c

Integrating 1.37896 c dV from 0.8 to 0.4 gives
-0.551582 c

If I use the original p (that I found) and v (given) for c,
-0.551582(p*v) = -0.551582(49860*0.8) = -22kJ, which isn't the answer. Where did I go wrong?

No.

We have $$PV^ \gamma = C ==> P = C/V^ \gamma$$

$$W = \int PdV = \int CV^- ^\gamma dV = \frac{C V^ 1 ^- ^\gamma}{1 - \gamma}$$

Oh, I understand, thanks for your help!

The solution for future reference:
p*0.8 = 15*8.31*320
p -> 49860 Pa (Initial pressure)

p*(0.8^1.44) = c (0.8 is the initial volume)
c -> 36157.8

w = Integrate[c (v^-1.44), dv]
w = -(2.27273 c)/v^0.44
-> wf - wi = -32327.8 J

## 1. What is an adiabatic process?

An adiabatic process is one in which there is no transfer of heat between the system and its surroundings. This means that the change in internal energy of the system is solely caused by work done on or by the system.

## 2. How is work done in an adiabatic process?

In an adiabatic process, work is done by the system if the volume of the system decreases while the pressure remains constant. Conversely, work is done on the system if the volume increases while the pressure remains constant.

## 3. What is the formula for work done in an adiabatic process?

The formula for work done in an adiabatic process is W = P(V2 - V1), where W is the work done, P is the pressure, V2 is the final volume, and V1 is the initial volume.

## 4. How does work done in an adiabatic process affect the temperature of the system?

In an adiabatic process, the temperature of the system changes due to the change in internal energy caused by work done. If work is done on the system, the temperature increases, and if work is done by the system, the temperature decreases.

## 5. Can work done in an adiabatic process be reversed?

No, work done in an adiabatic process cannot be reversed. In an adiabatic process, there is no transfer of heat, so the change in internal energy is solely caused by work. Reversing the work would require an equal and opposite amount of work to be done, but this would result in a transfer of heat, making it a non-adiabatic process.

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