Adiabatic compression of ideal diatomic gas

[Fizz_Geek]

Homework Statement

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

Homework Equations

PV^γ=P₀V₀^γ

Where γ = C_P/C_V

And for an ideal diatomic gas, we have:

γ=$\frac{7}{5}$

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:

PV^γ=P₀V₀^γ

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?

Thanks in advance.

 

[rude man]

I did not get your answer.

But if I may:

pV^γ = c = p₁V₁^γ

W = ∫pdV = ∫(c/V^γ)dV = c∫V^-γdV with your limits (I also got V₂ = 0.22).

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

 

[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!

 

[rude man]

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 ? Yes.

Then W = - 19 kJ ? Yes!

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!

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

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

 

[Fizz_Geek]

Oh! Thanks very much for your time!

 

[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

 

[rude man]

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

AM

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

Good thought!

 

[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.

 

[rude man]

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

Don't need it.

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

 

[Fizz_Geek]

Oh! Okay, thanks for all your help!