Work done pumping air into a bottle

[Funky_Sp]

Hi there, i am struggling with the following problem. Air is pumping into a bottle with volume V and pressure Pi until it reaches a final pressure P_f. The temperature remains the same during the process and the gas is an ideal one.
We have to calculate the work that is done.

I am not quite sure how deal with it, since the volume is constant i cannot use the trivial expression W=\int_{Vi}^{Vf}pdV
I tried to derive an equation for the work, depending on the change in density and number of particles but I don't have enough information for that. My next attempt was to use the general expression for the work, W=\int_{i}^{f}Fdx but after a little calculus I am coming back to the first equation. My last attempt was to proceed with that , and state, that since the gas is an ideal one then p_{i}V_{i}=p_{f}V_{f}\Rightarrow \frac{p_{i}}{p_{f}}=\frac{V_{i}}{V_{f}}\Rightarrow ln\frac{p_{i}}{p_{f}}=ln\frac{V_{i}}{V_{f}}
but I think this is silly.

Does anyone has an idea how to solve it?

 

[Bestfrog]

I think that you're missing the fact that the number of moles $n$ increases with the pressure. The only two variables that change (in the equation of ideal gasses $pV=nRT$) are $p$ and $n$.

 

[haruspex]

I think that you're missing the fact that the number of moles $n$ increases with the pressure. The only two variables that change (in the equation of ideal gasses $pV=nRT$) are $p$ and $n$.

Quite so, but isn't there missing information? What is the initial pressure of the air being pumped in?

 

[Funky_Sp]

Thanks for your answers, haruspex you are right there are a lot of missing information, so actually i am looking for a general expression for the work and not a number.
I know that \frac{p_{i}}{p_{f}}=\frac{n_{i}}{n_{f}} but that doesn't help me so much.
My next thought is, that dV is actually the volume from the air of the environment that is pumping in and not the volume of the gas in the bottle. But that doesn't work out.

 

[haruspex]

My next thought is, that dV is actually the volume from the air of the environment that is pumping in and not the volume of the gas in the bottle.

Seems to me there are two parcels of air that change volume, that which was already inside and that which was added.

 

[Funky_Sp]

Actually, I went on with that , and at the end I got something that seems right to me . The derivation is though quite long so I am posting only the final equation.

W=[p(ln(\frac{p_{f}}{p_{0}}-1)+p_{0})]V

p_f indicates the pressure in the bottle after the compression, p₀ the atmospheric pressure and V the volume of the bottle.

 

[Nidum]

The pumping work is done external to the bottle by whatever is doing the pumping . This pumping may be more or less efficient depending in reality on a number of factors but usually the pumping process is considered for simplicity to be either an adiabatic or an isothermal process .

Given the pressure and temperature of air in the bottle at any time during the process of pressurisation the mass of air contained and the energy stored in the air are determinate and can be found by simple calculation .

 

[Bestfrog]

I was thinking about the fact that you know the internal energy $U$ of the ideal gas before and after, so $L=\Delta U$.

 

[Nidum]

Yes you can work out the energy gained by the air . The pumping work though is something different .

 

[Bestfrog]

Yes you can work out the energy gained by the air . The pumping work though is something different .

So, how to find it?

 

[haruspex]

So, how to find it?

I'm unsure what interpretation @Nidum is suggesting, but here's one...
Since we don't know the original pressure of the added air, we can take each parcel of air to be injected to be already at the current pressure in the container. If a small number of moles, dn, is injected, there is a corresponding increase dP in pressure. You can write out the work done in that, and integrate.

 

[Bestfrog]

I'm unsure what interpretation @Nidum is suggesting, but here's one...
Since we don't know the original pressure of the added air, we can take each parcel of air to be injected to be already at the current pressure in the container. If a small number of moles, dn, is injected, there is a corresponding increase dP in pressure. You can write out the work done in that, and integrate.

I arrive to this$$\frac{dp}{p}=\frac{dn}{n}$$
So $p=n\cdot e^C$, and finding the integration constant I have $C=ln(RT/V)$, but I return to the initial ideal gasses equation

 

[Nidum]

See if you can solve this simple problem :

A cylinder is closed at one end and fitted with a frictionless piston . The piston is initially located at the opposite end of the cylinder from the closed end .

What is the work done when the piston is forced part way down the cylinder so that the initially large volume of air in the cylinder is compressed to a smaller volume ?

Assume that the piston only moves relatively slowly and consider two different types of compression process : (a) adiabatic and (b) isothermal .

 

[Chestermiller]

@Funky_Sp I'm just guessing, but would it be correct to say that you are currently learning about the open system (control volume) version of the first law of thermodynamics?

 

[Bestfrog]

See if you can solve this simple problem :

A cylinder is closed at one end and fitted with a frictionless piston . The piston is initially located at the opposite end of the cylinder from the closed end .

What is the work done when the piston is forced part way down the cylinder so that the initially large volume of air in the cylinder is compressed to a smaller volume ?

Assume that the piston only moves relatively slowly and consider two different types of compression process : (a) adiabatic and (b) isothermal .

Part (b) The work done is $W=nRT ln\frac{V_i}{V_f}$
Part (a) If i consider in my formula the change of volume (it seems to me that we know he final volume but not the final pressure), I have $W=\frac{3}{2}nRT_i \cdot ((\frac{V_f}{V_i})^{-\gamma +1} - 1)$ for a monoatomic ideal gas.

 

[Chestermiller]

Part (b) The work done is $W=nRT ln\frac{V_i}{V_f}$
Part (a) If i consider in my formula the change of volume (it seems to me that we know he final volume but not the final pressure), I have $W=\frac{3}{2}nRT_i \cdot ((\frac{V_f}{V_i})^{-\gamma +1} - 1)$ for a monoatomic ideal gas.

That's not what I get. Based on the open system (control volume) version of the first law, I obtain: $$W=V(P_f-P_i)$$

 

[Bestfrog]

That's not what I get. Based on the open system (control volume) version of the first law, I obtain: $$W=V(P_f-P_i)$$

I was answering to the "problem" of Nidum in #13

 

[Chestermiller]

I was answering to the "problem" of Nidum in #13

Oh, I missed that. Sorry.

My answer is for the original problem as stated.