- #1

Fizz_Geek

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## 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?

P

_{0}= 1.9 atm = 1.93e5 Pa

V

_{0}= 0.30 m

^{3}

P = 2.9 atm = 2.94e5 Pa

## Homework Equations

PV

^{γ}=P

_{0}V

_{0}

^{γ}

Where γ = C

_{P}/C

_{V}

And for an ideal diatomic gas, we have:

γ=[itex]\frac{7}{5}[/itex]

## The Attempt at a Solution

W = [itex]\int[/itex]P dV

= [itex]\int[/itex] P

_{0}(V

_{0}/V)

^{[itex]\frac{7}{5}[/itex]}dV

= P

_{0}V

_{0}

^{[itex]\frac{7}{5}[/itex]}(V

^{[itex]\frac{-2}{5}[/itex]}-V

_{0}

^{[itex]\frac{-2}{5}[/itex]})/([itex]\frac{-2}{5}[/itex])

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

PV

^{γ}=P

_{0}V

_{0}

^{γ}

Plugging in the values, I get:

V = 0.22 m

^{3}

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.