Calculate the final temperature of the helium

[Albert1017]

Homework Statement

A rigid cylinder contains helium (molar mass 4kg/kmol) at a pressure of 5 bar and a temperature of 15 oC. The cylinder is now connected to a large source of helium at 10 bar and 15 oC, and the valve connecting the cylinder is closed when the cylinder pressure has risen to 8 bar. Calculate the final temperature of the helium in the cylinder assuming that the heat transfer during the process is negligibly small. Take Cv for helium as 3.12 KJ/KgK

The Attempt at a Solution

I'm trying to combine the two pressure at phase 1 but in vain.
The answer is 61.5 oC

 

[member 392791]

where's your mass and energy balances??

 

[rude man]

Seems to me that, since there is no heat transferred and no work done, that the final temperature must also be 15C assuming an ideal gas since then dU = dQ - dW = 0 and for an ideal gas dU = function of T only.


The final no. of moles is then given by p_2 V = n_2 RT.

 

[Chestermiller]

There are two equivalent approaches that can be used to solve this problem. One is a closed system approach, and the other is an open system approach. Both approaches lead to the same equations and to the same final answer. In the closed system approach, we treat the system as the gas originally within the cylinder plus the gas that eventually enters the cylinder from the large source. In the open system approach, we consider the cylinder as an open system into which the outside gas flows.

I'm going to focus on the closed system approach, although, in the OP's course, I'm guessing that they must currently be studying the open system approach. Between the initial and final equilibrium states of the system, there is no heat transfer to the system, but there is work done by the gas within the large source on that portion of the gas within the large source that enters the cylinder (the latter is part of our closed system). The amount of work done in pushing the gas into the cylinder is p_sv_sn_s, where p_s is the pressure within the large source, v_s is the molar specific volume of the gas within the large source, and n_s is the total number of moles of gas that enter the cylinder from the large source. According to the first law, this work must be equal to the change in internal energy of the system:
ΔU=(n_s+n₀)C_vM(T-288) where n₀ is the number of moles originally in the cylinder, M is the molar mass of the helium, and T is the final absolute temperature. So, from the first law,

(n_s+n₀)C_vM(T-288)=p_sv_sn_s
This equation can be combined with the ideal gas law to determine all the unknowns.

So, OP, show us how you proceed.

Chet

 

[rude man]

Thanks for the correction Chet. I was very uneasy about this one, which is why I started with 'Seems to me ...". Good thing I did.

Hard for me even now to visualize the work done.

 

[Chestermiller]

Thanks for the correction Chet. I was very uneasy about this one, which is why I started with 'Seems to me ...". Good thing I did.

Hard for me even now to visualize the work done.

It might help to picture an invisible membrane (with zero structural rigidity) surrounding the gas that eventually enters the cylinder, and separating it from the bulk of the gas in the large source. The pressure of the surroundings acting on this membrane is the pressure within the large source (throughout the entire injection). The volume change of the system (consisting of this gas and the gas initially in the cylinder) is just the volume of the gas within the membrane to begin with.

Chet

 

[Chestermiller]

Albert1017, are you still out there? How sure are you about that 61.5 C? I get a different answer. I get 47 C. Can you provide a solution that gives 61.5 C?

Chet