# Adiabatic compression of ideal diatomic gas

1. Jan 26, 2013

### Fizz_Geek

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

An ideal diatomic gas, with rotation but no oscillation, undergoes an adiabatic compression. Its initial pressure and volume are 1.9 atm and 0.30 m3. It's final pressure is 2.9 atm. How much work is done by the gas?

P0 = 1.9 atm = 1.93e5 Pa
V0 = 0.30 m3
P = 2.9 atm = 2.94e5 Pa

2. Relevant equations

PVγ=P0V0γ

Where γ = CP/CV

And for an ideal diatomic gas, we have:

γ=$\frac{7}{5}$

3. The attempt at a solution

W = $\int$P dV
= $\int$ P0(V0/V)$\frac{7}{5}$ dV
= P0V0$\frac{7}{5}$(V$\frac{-2}{5}$-V0$\frac{-2}{5}$)/($\frac{-2}{5}$)

Then we can solve for V from the adiabatic condition, since we were given the final pressure:

PVγ=P0V0γ

Plugging in the values, I get:

V = 0.22 m3

Then, plugging this value into the work equation, I get:

W = -1.85e5 J or 185 kJ of work done by the gas.

Sorry about my notation, I'm new to LaTex. So, did I do this right?

2. Jan 26, 2013

### rude man

But if I may:

pVγ = c = p1V1γ

W = ∫pdV = ∫(c/Vγ)dV = c∫VdV with your limits (I also got V2 = 0.22).

Also, the work done BY the gas must be a negative number.

Last edited: Jan 26, 2013
3. Jan 26, 2013

### Fizz_Geek

That's what I was trying to do, but my calculator died, so I was using the graphing calculator app on my phone.

Did you get c = 3.577e4 ?

Then W = - 19 kJ ?

If not, can you just tell me if the method is correct or not? Did I use the right value of gamma and the right equations?

Thanks again!

4. Jan 26, 2013

### rude man

EDIT:
see above .
You did fine, it was my goof.

I didn't double-check you value of γ though.

Last edited: Jan 26, 2013
5. Jan 26, 2013

### Fizz_Geek

Oh! Thanks very much for your time!

6. Jan 26, 2013

### Andrew Mason

There is an easier way to do this using $TV^{(\gamma-1)} = PV^{\gamma}/nR = \text{constant}$ to find an expression for ΔT and then use $\Delta U = nC_v\Delta T = W$ (adiabatic) to find W.

AM

Last edited: Jan 26, 2013
7. Jan 26, 2013

### rude man

I didn't think ΔU = CVdT was covered in an intro physics course, then I found it in my Resnick & Halliday ...

Good thought!

Last edited: Jan 26, 2013
8. Jan 26, 2013

### Fizz_Geek

Actually, that formula is in my intro book. But how would I use it here? I don't have the value of n.

9. Jan 26, 2013

### rude man

Don't need it.

T1 = p1V1/nR
T2 = p2V2/nR
ΔT = T2 - T1 = (1/n)(p2V2 - p1V1)/R
ΔU = ncvΔT = W; the n 's cancel out.
cv is the molar specific heat at const. volume. R is the universal gas constant.

Last edited: Jan 26, 2013
10. Jan 26, 2013

### Fizz_Geek

Oh! Okay, thanks for all your help!