B Irreversible expansion of gas against gas

[Christofer Br]

TL;DR Summary

If I expand a gas on one side of the piston to compress a gas on the other, will they end up at the same temperature if the piston and the cyllinder are perfect insulators?

Let's say I have a liter of gas at pressure of 4 atm and T=900K. I use it to move perfect massless frictionless and insulative piston to compress a liter of 1 atm, T=300K gas. When the pressure on both sides is equal and the piston stops moving, will the temperature on both sides of the piston be the same?

 

[mfig]

Use this isentropic relation for each gas (a and b) , setting $T_{2a} =T_{2b}$ and similarly for the final pressures. See what relation must obtain between initial quantities in order for that to hold.

$\frac{T_2}{T_1}= \left( \frac{P_2}{P_1}\right) ^{\frac{\gamma-1}{\gamma}}$

 

[Chestermiller]

Use this isentropic relation for each gas (a and b) , setting $T_{2a} =T_{2b}$ and similarly for the final pressures. See what relation must obtain between initial quantities in order for that to hold.

$\frac{T_2}{T_1}= \left( \frac{P_2}{P_1}\right) ^{\frac{\gamma-1}{\gamma}}$

The equation that you suggest is not valid for the irreversible change that occurs.

 

[Chestermiller]

Summary:: If I expand a gas on one side of the piston to compress a gas on the other, will they end up at the same temperature if the piston and the cyllinder are perfect insulators?

If the piston and cylinder are perfect insulators, you would not expect the two gases to end up at the same temperature. For an irreversible process like this one, the most you can say using thermodynamics is that the total change in internal energy of the two gases is zero, and the final pressures are equal.

 

[Chestermiller]

Irrespective of the initial temperatures, I can prove that, if the two initial volumes are equal, the final pressure will be the arithmetic average of the two initial pressures. To solve for the remainder of the final conditions (temperatures and volumes), it would be necessary to include a description of how the force per unit area on the piston faces varies during the process as a function of the volume change. One approximation that might be pretty accurate might be to assume that the force per unit area remains constant throughout the process at the final pressure. This would certainly entail an increase in the entropy of the combination of the gases, in line with the irreversibility of the process. If anyone is interested, I can provide the results of the calculations for this approximation.

 

[Christofer Br]

Irrespective of the initial temperatures, I can prove that, if the two initial volumes are equal, the final pressure will be the arithmetic average of the two initial pressures. To solve for the remainder of the final conditions (temperatures and volumes), it would be necessary to include a description of how the force per unit area on the piston faces varies during the process as a function of the volume change. One approximation that might be pretty accurate might be to assume that the force per unit area remains constant throughout the process at the final pressure. This would certainly entail an increase in the entropy of the combination of the gases, in line with the irreversibility of the process. If anyone is interested, I can provide the results of the calculations for this approximation.

Of course I am interested!

 

[Chestermiller]

Of course I am interested!

If we apply the first law of thermodynamics to each of the gases, we have $$\Delta U_1=n_1C_v(T_1-T_{10})=-W$$and$$\Delta U_2=n_2C_v(T_2-T_{20})=+W$$where W is the work done by gas 1 on gas 2. Using the ideal gas law to substitute for the temperatures, we have $$\Delta U_1=\frac{C_v}{R}[P(V_0+\Delta V)-P_{10}V_0]=-W$$and$$\Delta U_2=\frac{C_v}{R}[P(V_0-\Delta V)-P_{20}V_0]=+W$$where P is the final pressure at equilibrium, $V_0$ is the initial volume of each of the two gases, and $\Delta V$ is the increase in volume of gas 1. If we add these two equations together, we obtain: $$\frac{C_v}{R}[2PV_0-(P_{10}+P_{20})V_0]=0$$Solving this equation for the final pressure P, we obtain: $$P=\frac{(P_{10}+P_{20})}{2}$$So, if the two initial volumes are equal, the final pressure is the arithmetic mean of the initial pressures of the two gases.

OK so far?

 

[Christofer Br]

Thank you, this is very useful (and clear so far).

 

[Chestermiller]

Under the assumption that, during the expansion, the force per unit area of each gas on the piston is constant and equal to the final pressure $P=\frac{(P_{10}+P_{20})}{2}$, the first law applied to gas 1 becomes:
$$\Delta U_1=n_1C_v(T_1-T_{10})=-P(V_1-V_{0})$$where $V_1$ is the final volume of gas 1. Applying the ideal gas law to the initial and final volumes, we have:$$n_1C_v(T_1-T_{10})=-P\left(\frac{n_1RT_1}{P}-\frac{n_1RT_{10}}{P_{10}}\right)$$or, equivalently, $$(C_v+R)T_1=\left(C_v+R\frac{P}{P_{10}}\right)T_{10}$$So, solving for the final temperature of gas 1, we obtain:
$$T_1=\frac{\left[1+(\gamma-1)\frac{P}{P_{10}}\right]}{\gamma}T_{10}$$where $\gamma=C_p/C_v$. Similarly, the final temperature of gas 2 is given by: $$T_2=\frac{\left[1+(\gamma-1)\frac{P}{P_{20}}\right]}{\gamma}T_{20}$$

So, now we know the final pressures and final temperatures of the gases. From the ideal gas law, we can get the final volumes. So, what values do you calculate for these parameters in the problem you specified?

 

[Christofer Br]

Actually I have a problem with calculating the final pressure when the initial volumes aren't equal; I end up with:

(Cv/R)*[P*V_20+P*V_10-(V_10*P_10-V_20*P_20)]=0

when P=10, P_10=100, P_20=1, V_20=1 I calculate V_10 (initial volume of the gas doing the work) as 1/10

When I plug in the numbers back into the equation for U1, i find that the work done by expanding gas is zero, which is impossible. Why did I do wrong?

 

[Chestermiller]

Actually I have a problem with calculating the final pressure when the initial volumes aren't equal; I end up with:

(Cv/R)*[P*V_20+P*V_10-(V_10*P_10-V_20*P_20)]=0

when P=10, P_10=100, P_20=1, V_20=1 I calculate V_10 (initial volume of the gas doing the work) as 1/10

When I plug in the numbers back into the equation for U1, i find that the work done by expanding gas is zero, which is impossible. Why did I do wrong?

I can't read what you wrote. When the initial volumes are not equal, I get:
$$\Delta U_1=\frac{C_v}{R}[P(V_{10}+\Delta V)-P_{10}V_{10}]=-W$$and$$\Delta U_2=\frac{C_v}{R}[P(V_{20}-\Delta V)-P_{20}V_{20}]=+W$$
So, $$P=P_{10}\frac{V_{10}}{V_{10}+V_{20}}+P_{20}\frac{V_{10}}{V_{10}+V_{20}}$$In your post, you gave values for $P_{10}=100$, $P_{20}=1$, and $V_{20}=1$, but I didn't see a specified value of $V_{10}$.