How Does a Portable Device for Pressurizing Tires Impact Thermodynamics?

[siddharth]

The question verbatim is:

"A portable device for pressurising tyres consists of a small container of 1.5 cft volume, initially filled with, say, CH₄ at 500 psia. When a tyre (volume=0.5 cft) is to be filled, the gas is allowed to flow into it from the container through a valve till the pressure in the tyre reaches 50 psia. The process is essentially adiabatic and the initial pressure in the tyre may be taken as 15 psia. Let the subscripts 1 & 2 and i & f refer to the initial and final states of the gas in the container and tyre respectively. Assume that P1=500 psia, Pi=15, T1=Ti=80 F

a) Assume that the filling process is reversible. Use the chart to read v,s and h for the gas (i)initially in the container (ii) initially in the tyre and (iii) finally in the tyre."
b) Calculate m1, mi and mf
c) What's the value of s2? Read t2 and h2 from the chart
d) Use the first law for container + tyre to find m2(h2-p2v2)
e) Solve for P2 () by trial and error so as to satisfy the first law
f) Calculate the change in entropy of
(i)The closed system (container + tyre)
(ii)The environmentFirst of all, I think there's some ambiguity in the statement of the question on the notations. Leaving that aside, I have a doubt regarding the final state of the air in the tyre. The thing is that T_f for the tyre isn't known. So how can I calculate T_f (so that I can then calculate h,v,s from the chart)?
Now the process is adiabatic and reversible, so the entropy change for the whole system (ie, container + tyre, neglect valve) is 0. But, that doesn't help me find T_f because the entropy change in the tyre alone wouldn't be 0. I think there's some information which needs to be given, or some assumption to make to calculate the final state. Once I do that, the rest of the problem seems easy.

 

[Valkarie]

In order to calculate T_f, you need to make some assumptions about the process. For example, you can assume that the temperature of the air in the tyre remains constant during the filling process. This means that you can use the equation:P_2V_2 = P_1V_1 to calculate P_2 and then use the ideal gas law to calculate T_2. Once you have T_2, you can then find h_2, v_2 and s_2 from the chart. For part (f), the change in entropy of the closed system (container + tyre) is equal to the change in entropy of the environment, since the process is reversible. So the entropy change of the environment would be 0 as well.

 

[piercas]

Thank you for your question. In this scenario, it seems that the final state of the air in the tyre is not explicitly given. However, based on the information provided, we can make some assumptions to solve the problem.

First, let's assume that the tyre is initially at the same temperature as the container (80°F). This means that the initial and final temperatures of the air in the tyre are the same. We can then use the ideal gas law (P1V1/T1 = P2V2/T2) to solve for the final pressure (P2) in the tyre.

Next, we can use the ideal gas law again to calculate the initial and final masses of the gas in the container and tyre. We know the initial volume and pressure of the gas in the container and the final pressure and volume of the gas in the tyre. This allows us to calculate the initial and final masses (m1 and mf) using the ideal gas law (m = PV/RT).

Once we have the final pressure and mass of the gas in the tyre, we can use the chart to read the specific entropy and enthalpy values for the gas in the tyre. This will allow us to calculate the change in entropy and enthalpy for the gas in the tyre.

For the closed system (container + tyre), we can use the first law of thermodynamics (ΔU = Q - W) to calculate the change in internal energy. Since the process is adiabatic, there is no heat transfer (Q=0). Also, since the container is assumed to be insulated, there is no work done (W=0). This means that the change in internal energy is also 0.

Finally, we can use the definition of entropy (ΔS = Q/T) to calculate the change in entropy for the environment. Since the process is adiabatic, there is no heat transfer (Q=0). We also know the initial and final temperatures of the environment (80°F and 50°F, respectively), so we can calculate the change in entropy.

I hope this helps clarify the steps needed to solve this problem. Please let me know if you have any further questions.