Thermo: A Flat Tire and a Pressurized Tank

[Thyferra2680]

Homework Statement

We have a tank of air at 7 atm, 30 L, and 293 Kelvin(20 Celsius), and a flat tire with 1atm, 10 L (Constant), also at 293 K (20 C). The valve is opened quickly and closed quickly when the pressure in the tank reaches 6 atm. Using the Ideal gas assumption and taking the heat capacity of the air as Cp = 3.5 R...

Determine: the temperature of the air in the tire and the tank immediately after filling, and the eventual pressure in the tank and tire after equilibration with room temperature.

Homework Equations

PV = RT

T2/T1 = (V2/V1)^ (1-$$\gamma$$)

dU = dW (Adiabatic)

dQ = 0 (Adiabatic)

dW = - $$\int$$PdV

Cv = Cp-R

$$\gamma$$ = Cp/Cv = 1.4

The Attempt at a Solution

The first thing I did was assume that the problem was a closed, adiabatic system, with a constant volume. For the first part, I assumed that "immediately after filling" meant the split second you close the tank, implying that each portion of the system is "open" to each other. Intuitively, I also know that a tire tends to warm up when air is pumped in, so I also hoped that my answer might coincide with that conclusion.

However I looked at the tank first. I said that T2/T1 = (V2/V1)^ (1-$$\gamma$$), and plugged these sets of numbers in:

T2/293K = (40L/30L)^(-.4)
T2(tank) = 261.15K, -11.84 C (in the tank)

T2/293K = (40L/10L)^(-.4)
T2 (tire) = 168.28 K, -104.71 C (in the tire)

This, of course goes against my initial belief that a tire usually warms slightly when you compress it. Is my process for the first part correct?

 

[Chestermiller]

This is not done correctly. The gas within the tank at any time has experienced an adiabatic reversible expansion to force the gas ahead of it out of the tank (through the valve). So the temperature and pressure in the tank are related by $$\frac{T_2}{T_1}=\left(\frac{P_2}{P_1}\right)^{(\gamma-1)/\gamma}$$So, $$T_2=293\left(\frac{6}{7}\right)^{0.4/1.4}=280.4$$

The initial number of moles in the tank is $$n_1=\frac{P_1V}{RT_1}=\frac{(7)(30)}{(0.082)(293}=8.74$$
The final number of moles in the tank is $$n_2=\frac{P_2V}{RT_2}=\frac{(6)(30)}{(0.082)(280.4)}=7.83$$

So the number of moles of air leaving the tank and going into the tire while the valve is open is $$\Delta n=8.74-7.83=1.01$$The initial number of moles in the tire is $$n_i=\frac{P_iV_{tire}}{RT_i}=\frac{(1)(10)}{(0.082)(293)}=0.42$$ So the final number of moles of air in the tire is $$n_f=0.42+1.01=1.43$$

During the filling operation, the process is adiabatic for the combination of tire and tank, and no work is done on this combination. So the change internal energy of the combined system of tire and tank is zero. So, $$n_1C_v(T_1-T_{ref})+n_iC_v(T_i-T_{ref})=n_2C_v(T_2-T_{ref})+n_fC_v(T-T_{ref})$$where $T_{ref}$ is an arbitrary reference temperature which cancels from the equation. From this, it follows that $$(7.83)(280.4-293)+1.43(T_f-293)=0$$or$$T_f=362\ K$$

The final pressures in the tank and tire after they equilibrate with room temperature is determined by the ideal gas law.