# Adiabatic compression of ideal diatomic gas

• Fizz_Geek
In summary, an ideal diatomic gas undergoes an adiabatic compression, resulting in a change in pressure and temperature. Work is done by the gas in the form of a negative number.
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

PVγ=P0V0γ

Where γ = CP/CV

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:

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?

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:
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!

Fizz_Geek said:
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.

Last edited:
Oh! Thanks very much for your time!

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:
Andrew Mason said:
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 = CVdT was covered in an intro physics course, then I found it in my Resnick & Halliday ...

Good thought!

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

Fizz_Geek said:
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 = ncvΔT = W; the n 's cancel out.
cv is the molar specific heat at const. volume. R is the universal gas constant.

Last edited:
Oh! Okay, thanks for all your help!

## 1. What is adiabatic compression?

Adiabatic compression is the process of compressing a gas without any heat entering or leaving the system. This means the temperature of the gas remains constant during the compression.

## 2. What is an ideal diatomic gas?

An ideal diatomic gas is a type of gas composed of molecules with two atoms, such as oxygen (O2) or nitrogen (N2). It is assumed to follow the ideal gas law, which describes the relationship between pressure, volume, and temperature.

## 3. How does adiabatic compression affect an ideal diatomic gas?

During adiabatic compression, the volume of an ideal diatomic gas decreases, causing the pressure to increase. This is because the energy of the gas molecules is concentrated in a smaller space, resulting in more frequent collisions and higher pressure.

## 4. What is the equation for adiabatic compression of ideal diatomic gas?

The equation for adiabatic compression of an ideal diatomic gas is P1V1γ = P2V2γ, where P1 and V1 are the initial pressure and volume of the gas, P2 and V2 are the final pressure and volume, and γ is the adiabatic index, which is a constant value depending on the gas.

## 5. What factors can affect the adiabatic compression of an ideal diatomic gas?

The adiabatic compression of an ideal diatomic gas can be affected by the initial temperature, pressure, volume, and the adiabatic index of the gas. The type of gas and the conditions of the surroundings can also play a role in the compression process.

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