Instantaneous Gas Compression: temperature increase?

[FranzS]

If I a have a gas confined in a certain initial volume Vin at a certain pressure Pin and at a certain temperature Tin, and istantaneously compress it down to a final volume Vfin < Vin, how do I calculate the increase in temperature?
Assume I know the exact pressure curve (P vs. V).
The system cannot exchange heat (I guess that's redundant with "istantaneous compression").

 

[Dale]

That is called adiabatic compression. I can look up the formula later, but if you need it quickly just search for adiabatic compression.

 

[etotheipi]

Do you mean, like, suddenly increasing the pressure of the surroundings from $p_i$ to $p_f$ so that the gas undergoes irreversible and adiabatic compression?

 

[Arjan82]

You can do that with the Poisson equations. The limitation is that the system must be in thermodynamic equilibrium at all times, so not 'instantaneous' compression.

 

[Arjan82]

The Poisson equations are derived for a barotropic flow and calorically perfect gas.

 

[etotheipi]

@Arjan82 and @Dale your equations only apply to reversible processes but to me at least it doesn't sound from OPs description (instantaneous $\leftrightarrow$ fast?) like a reversible process. @FranzS, can you clarify?

I had in mind that if - for example - the surrounding pressure is raised suddenly from $p_i$ to $p_f > p_i$ and held at $p_f$ then the temperature change due to the irreversible compression, given $\partial U / \partial T = C_V$ always for an ideal gas and also $Q = 0$, would satisfy $\Delta T = -\frac{p_f}{C_V}\Delta V$

 

[etotheipi]

If there is no time for heat exchange then entropy is constant and the process is indeed reversible.

It's not necessarily true i.e. you can have irreversible adiabatic processes ($Q=0$). For an irreversible process $dS > \delta Q / T$ so even with $\delta Q = 0$ the system undergoes an entropy increase; it reflects that during an irreversible process entropy is generated inside the system

e.g. my example in #7 is adiabatic, irreversible and $\Delta S > 0$

 

[Chestermiller]

@Arjan82 and @Dale your equations only apply to reversible processes but to me at least it doesn't sound from OPs description (instantaneous $\leftrightarrow$ fast?) like a reversible process. @FranzS, can you clarify?

I had in mind that if - for example - the surrounding pressure is raised suddenly from $p_i$ to $p_f > p_i$ and held at $p_f$ then the temperature change due to the irreversible compression, given $\partial U / \partial T = C_V$ always for an ideal gas and also $Q = 0$, would satisfy $\Delta T = -\frac{p_f}{C_V}\Delta V$

Yes, as you said, this process definitely results in an increase in entropy. Nice assessment!

 

[FranzS]

Do you mean, like, suddenly increasing the pressure of the surroundings from $p_i$ to $p_f$ so that the gas undergoes irreversible and adiabatic compression?

Yes but I'm confused.
I'd like to clarify my initial statement:

Assume I know the exact pressure curve (P vs. V).

I meant I know the isothermal $P$ vs. $V$ curve for any given temperature.
But this compression is not isothermal of course.
And I do not know the final pressure $P_f$, actually that's the ultimate goal which I'd like to calculate from the increase in temperature!

 

[FranzS]

to me at least it doesn't sound from OPs description (instantaneous $\leftrightarrow$ fast?) like a reversible process. @FranzS, can you clarify?

For "istantaneous" I mean that it occurs in a time $t \rightarrow 0$, which I believe excludes the possibility of heat exchange with the surroundings.
Please note that I know exact initial volume, exact final volume, exact initial pressure. I do not know final pressure in this process, although I know exact final pressure*** if it were an isothermal process (in other words, a compression that occurs in a time $t \rightarrow \infty$ where the system can this time exchange heat with the surroundings).
*** With exact [isothermal] final pressure, I mean compressibility factor included.

 

[etotheipi]

And I do not know the final pressure $P_f$, actually that's the ultimate goal which I'd like to calculate from the increase in temperature!

Okay, well I suppose we ought to start with $T_f - T_i = -\frac{p_f}{C_V}(V_f - V_i)$ from post #6. For an ideal gas the equation $V = nRT/p$ applies at the initial and final equilibrium states so you may write $V_i = nRT_i/p_i$ and $V_f = nRT_f/p_f$, so$$T_f - T_i = \frac{-nRp_f}{C_V} \left( \frac{T_f}{p_f} - \frac{T_i}{p_i} \right) =\frac{nRT_i }{C_V}\left( \frac{p_f}{p_i} \right) - \frac{nRT_f}{C_V} $$I'll leave it to you to rearrange it how you like!

 

[FranzS]

Okay, well I suppose we ought to start with $T_f - T_i = -\frac{p_f}{C_V}(V_f - V_i)$ from post #6. For an ideal gas the equation $V = nRT/p$ is always true, so you may write $V_i = nRT_i/p_i$ and $V_f = nRT_f/p_f$, so$$T_f - T_i = \frac{-nRp_f}{C_V} \left( \frac{T_f}{p_f} - \frac{T_i}{p_i} \right) =\frac{nRT_i }{C_V}\left( \frac{p_f}{p_i} \right) - \frac{nRT_f}{C_V} $$I'll leave it to you to rearrange it how you like!

The problem is I would like to find out $\Delta T$ and use it to find $P_f$. But I guess I can't, since the two are related because of the work done during compression depends on $P$.
I'm using the real gas law $PV=ZnRT$ where I know a pretty accurate expression for the compressibility factor $Z(P, T)$.
Should I throw $PV=Z(P,T)nRT$ in your last equation? I guess it'll become a mess.

 

[etotheipi]

I'm not sure. Our resident thermodynamics expert @Chestermiller will probably know!

 

[Chestermiller]

The problem is I would like to find out $\Delta T$ and use it to find $P_f$. But I guess I can't, since the two are related because of the work done during compression depends on $P$.
I'm using the real gas law $PV=ZnRT$ where I know a pretty accurate expression for the compressibility factor $Z(P, T)$.
Should I throw $PV=Z(P,T)nRT$ in your last equation? I guess it'll become a mess.

The ideal gas law or any real gas equation of state is valid only at thermodynamic equilibrium. The rapid adiabatic compression of a gas is an irreversible process in which all the states that the gas passes through between the initial and final states are not thermodynamic equilibrium states. Therefore, the equation of state can not be used for these non-equilibrium states, and, if you try to use it, it will give the wrong answer for gas pressure on the piston face and for the work. In such a case, the only way that thermodynamics can be used to solve such a compression problem is to specify the pressure manually, as the pressure exerted by the piston face on the gas as a function of the gas volume. This is what etotheipi did, by specifying that the pressure on the gas is held constant during the compression at some specified value much higher than the original equilibrium gas pressure. This is about the best you are going to do.