Is this Gas Expansion Problem a Quasi-Static Adiabatic Process?

[Terry Bing]

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@Chestermiller
I am still having trouble figuring out if a given process is Quasi-static or not.
Consider the following case. The cylinder consists of an ideal gas at the bottom and a liquid of density ρ at the top separated by a piston of mass m and area A. Atmospheric pressure is p₀. Again, piston and cylinder are insulating. Initially, the piston is at equilibrium. A heater is turned on in the gas. The gas will expand causing the liquid to spill out. Is this process a quasi-static adiabatic process? In this case there are no sudden changes.

If so then the final temperature T₂(when piston reaches the top) should satisfy
T₂V₂^γ-1=T₁V₁^γ-1.
For a monatomic gas, we have γ=5/2, which gives us T₂=(0.5)^3/2T₁, [Since V₂=2V₁]
Is this correct?

 

[Chestermiller]

MODERATOR'S NOTE: Moved from other forum, so no template

@Chestermiller
I am still having trouble figuring out if a given process is Quasi-static or not.
Consider the following case. The cylinder consists of an ideal gas at the bottom and a liquid of density ρ at the top separated by a piston of mass m and area A. Atmospheric pressure is p₀. Again, piston and cylinder are insulating. Initially, the piston is at equilibrium. A heater is turned on in the gas. The gas will expand causing the liquid to spill out. Is this process a quasi-static adiabatic process?

It is not adiabatic because heat is being added to the gas. If the heat is added gradually enough, the process is quasi static.

In this case there are no sudden changes.
View attachment 220845
If so then the final temperature T₂(when piston reaches the top) should satisfy
T₂V₂^γ-1=T₁V₁^γ-1.

No. As I said, the expansion is not adiabatic, and this equation applies to an adiabatic reversible process.

For a monatomic gas, we have γ=5/2,

No. For a monatomic gas, $\gamma = 5/3$

Is this correct?

No.

In this problem, you need to start out by determining the pressure as a function of the change in piston elevation x (so that the work can be determined and the final temperature can be determined). What is the pressure on the gas initially? What is the pressure on the gas after the piston has risen by a distance x? What is the pressure on the gas when the piston has risen a distance h?

 

[Terry Bing]

In this problem, you need to start out by determining the pressure as a function of the change in piston elevation x (so that the work can be determined and the final temperature can be determined). What is the pressure on the gas initially? What is the pressure on the gas after the piston has risen by a distance x? What is the pressure on the gas when the piston has risen a distance h?

Thanks, I understand. I got half the answer. A number of hasty mistakes on my part in the attempt above. The process isn't adiabatic. It isn't given whether the gas is mono or diatomic, so we don't know γ.
At any instant of time, external pressure on the gas is P_ext=P₀+ρgy+mg/A, where y is the depth of the water layer. Work done by the gas
$$W=\int_{0}^{h} P_{ext} A dy= \int_{0}^{h} (P_0 A+\rho g y A+mg) dy=P_0 A h+\frac{1}{2} \rho g h^2 A+m g h$$
Now If we know the heat supplied,
$$Q-W=\Delta U$$
But Q is not given in the question. Also, we don't know γ to get ΔU in terms of T.

 

[Terry Bing]

I got it. We can use equation of state for an ideal gas. Assuming the process is quasi-static, both initial and final states will be equilibrium states. Then
$$P_1 V_1 / T_1=P_2 V_2/ T_2 \\ \implies T_2=( \frac{V_2}{V_1}) (\frac{P_2}{P_1}) T_1 \\ \implies T_2=2(\frac{p_0 A+m g}{p_0 A+\rho g h A+m g}) T_1$$

 

[Chestermiller]

I got it. We can use equation of state for an ideal gas. Assuming the process is quasi-static, both initial and final states will be equilibrium states. Then
$$P_1 V_1 / T_1=P_2 V_2/ T_2 \\ \implies T_2=( \frac{V_2}{V_1}) (\frac{P_2}{P_1}) T_1 \\ \implies T_2=2(\frac{p_0 A+m g}{p_0 A+\rho g h A+m g}) T_1$$

Very Good. It's just $\Delta U=\frac{C_v}{R}\Delta (PV)$