# Process for a monatomic gas in a cylinder

I am trying to gain an conceptual understanding of this problem:

Consider a piston dividng horizontal cylinder into two equal parts. No heat can flow from the piston or through the cylinder (the system is closed). Each part of the cylinder contains a monatomic gas of volume $V_0$, temperature $T_0$ and pressure $p_0$. Now from the inside of the left part of the cylinder, heat is added slowly until the pressure in the right part of the cylinder is $p_R$.

The problem is to find the temperature in the left part when the pressure in the right part is $p_R$. Therefore I am trying to understand the thermodynamic process taking place in the left part.

The first argument I thought of was that during the expansion the pressure on both sides of the piston are more or less equal since the heat is added slowly. So since the right part is compressed adiabatically, but that can't be right because there is heat in the system.

Then I thought (intuitively) that since the heat is added slowly, all of the heat must go into moving the piston thus giving $\Delta U = 0$ (isothermal process).

However this is only my intuition and I am unsure about it. My question is: How can I know that some of the added heat won't go into increasing the temperature of the gas?

The first argument I thought of was that during the expansion the pressure on both sides of the piston are more or less equal since the heat is added slowly.

That is correct.

So since the right part is compressed adiabatically, but that can't be right because there is heat in the system.

The right half is compressed adiabatically since heat cannot get into or out of that half, eg by conduction through the piston. I don't understand why you say "that can't be right because there is heat in the system." Heat does enter at the left side but none is conducted to the right.

Then I thought (intuitively) that since the heat is added slowly, all of the heat must go into moving the piston thus giving $\Delta U=0$ (isothermal process)... How can I know that some of the added heat won't go into increasing the temperature of the gas?

No, that does not follow. Heat added on the left both increases the temperature of the gas and does work by moving the piston. It is not an isothermal process. An increase in temperature is needed to increase pressure and move the piston to the right.

The gas on the right is compressed adiabatically so
$$p_R V_R^\gamma = p_0 V_0^\gamma$$ At all times we have $p_R=p_L$ because the piston is in quasi-static equilibrium, and $V_L+V_R=2V_0$ because the total volume is constant.

On the left we are increasing the temperature of the gas while keeping the mass constant so
$$\frac{p_L V_L}{T_L} = \frac{p_0 V_0}{T_0}$$

You will need to use these equations to eliminate $p_L, V_L, V_R$. For a monatomic gas $\gamma=\frac53$.

• Thank you for your answer. What I meant instead of "that can't be right because there is heat in the system" was that I thought that since the right part is adiabatically compressed and the pressure on both sides are equal the left part would be adiabatically expanded, but that is of course nonsense since there is heat added to the left part. I can see how one can find the temperature given your information, but now I'm thinking: can we even say something about which thermodynamic process is taking place in the left part? Commented Jul 11, 2016 at 15:58
• Ziko : I don't think there is any name for the process on the left. It is not adiabatic. Nor isothermal, nor iso-volumetric, nor isobaric (constant pressure). All of p, V and T are changing. Commented Jul 11, 2016 at 17:00