Thermodynamic pressure/temperature question

[physstudent1]

Homework Statement

Consider filling a cylinder of compressed argon from a high-pressure supply line. Before filling, the cylinder contains 10 bar of argon at room temperature the valve is then opened, exposing the tank to a 50 bar line at room temperature until the cylinder reaches 50bar. The valve is then closed. For argon take C_p =5/2 *R and the molecular weight to be 40kg/mol. You may use the ideal gas model.
a) what is the temperature after the valve is closed.
b)if the cylinder sits in storage for a long time how much heat is transferred.
c) what is the pressure of the cylinder when it is shipped (after sitting a long time)

Homework Equations

The Attempt at a Solution

Ok, so I'm not really worried about b or c yet since I can't get a.

Anyway My attempt:

I used an equation my prof gave us T2=(P2*(Cp/Cv)*T1)/(P2-P1 + P1/T1*(Cp/Cv)*T1

I plugged in the given value for Cp used Cv=Cp-R
used P2=50bar P1=10bar T1=298K and got 59.4 However I am pretty sure the temperature wouldn't go down, if anything it should go up since it is being compressed to a higher pressure. can anyone help maybe this equation is wrong or something? my prof writes really sloppy and I can't read it properly sometimes, let me know if this is the right forum too I wasn't sure where to put this question!

 

[Andrew Mason]

Homework Statement

Consider filling a cylinder of compressed argon from a high-pressure supply line. Before filling, the cylinder contains 10 bar of argon at room temperature the valve is then opened, exposing the tank to a 50 bar line at room temperature until the cylinder reaches 50bar. The valve is then closed. For argon take C_p =5/2 *R and the molecular weight to be 40kg/mol. You may use the ideal gas model.
a) what is the temperature after the valve is closed.
b)if the cylinder sits in storage for a long time how much heat is transferred.
c) what is the pressure of the cylinder when it is shipped (after sitting a long time)

Think of the gas in the cylinder being compressed by the inflowing gas (eg. think of the argon in the cylinder being contained in a sealed rubber balloon that is then compressed to 50 bar). The inflowing gas, however, does not experience a net compression.

Work out the temperature for an adiabatic compression of the gas that is originally in the cylinder. Then let the two mix so that the temperature evens out.

AM

 

[nlightner]

Hello,

Thank you for sharing your attempt at solving this problem. I can see that you have used the correct equation for finding the final temperature after compressing the gas. However, there may be a mistake in the values you have plugged in. The temperature should indeed go up after compressing the gas, as you mentioned. Here is how I would solve this problem:

a) To find the final temperature, we can use the ideal gas law: PV = nRT. The number of moles of argon can be calculated using the molecular weight (40kg/mol) and the given mass of argon in the cylinder. Therefore, n = (10 bar)(V) / (40 kg/mol)(0.0821 L*atm/mol*K)(298 K) = 0.0125 moles.

Now, using the ideal gas law again for the final state (50 bar and unknown temperature), we can solve for T:

(50 bar)(V) = (0.0125 moles)(0.0821 L*atm/mol*K)(T)

Solving for T, we get T = 894 K.

b) To find the amount of heat transferred during storage, we can use the first law of thermodynamics: ΔU = Q - W. Since the cylinder is in equilibrium, there is no change in internal energy (ΔU = 0). Therefore, the heat transferred (Q) is equal to the work done (W) on the gas. We can calculate the work done using the equation W = PΔV. The initial volume (V1) can be calculated using the ideal gas law (V1 = nRT1 / P1), and the final volume (V2) can be calculated using the same equation but with the final pressure (P2). Therefore, the work done is:

W = (50 bar)(V2 - V1) = (50 bar)(nRT2 / P2 - nRT1 / P1) = 0.0125 moles)(0.0821 L*atm/mol*K)(894 K - 298 K) = 47.2 L*atm.

c) Finally, to find the pressure of the cylinder when it is shipped after sitting for a long time, we can use the ideal gas law again. The temperature will remain the same (894 K) and the volume will also remain the same (since the cylinder