What is the different between filled rapidly and filled slowly?

[rei]

I don't understand this problem. What is the different between filled rapidly and filled slowly? I think in c, when it's filled slowly, the temperature would stay the same at T= 280 K but what's about a and b. How can I do those? Any hint, please. Oh, and here is the problem

Oxygen gas at T- 280 K and atmospheric pressure is used to fill in an oxygen tank in Cesna 210 aircraft. The oxygen cylinder has a volumn of 11.4 liters and the oxygen pressure when the tank is full is at 122 atmosphere.

If the tank is filled rapidly, the oxygen is compressed adiabatically. In this case,
a, how much oxygen fills the tank?
b, what is the final temperature of oxygen at the instant the tank is filled
c, If the tank is filled slowly and isothermally, how much oxygen (in kg) is required to fill the tank?

Thanks a lot!

 

[dextercioby]

I don't understand this problem. What is the different between filled rapidly and filled slowly? I think in c, when it's filled slowly, the temperature would stay the same at T= 280 K but what's about a and b. How can I do those? Any hint, please. Oh, and here is the problem

Oxygen gas at T- 280 K and atmospheric pressure is used to fill in an oxygen tank in Cesna 210 aircraft. The oxygen cylinder has a volumn of 11.4 liters and the oxygen pressure when the tank is full is at 122 atmosphere.

If the tank is filled rapidly, the oxygen is compressed adiabatically. In this case,
a, how much oxygen fills the tank?
b, what is the final temperature of oxygen at the instant the tank is filled
c, If the tank is filled slowly and isothermally, how much oxygen (in kg) is required to fill the tank?

Thanks a lot!

I think the answer to your question is contained in the problem's text,and u seemed to get it.If that tank is filled quickly,then the gas will suffer a general transformation in which it cannot ("it doesn't have time") change het with the environment.When it's filled slowly,then it is always in thermal equilibrium with the surroundings,and to do that,it changes heat constantly with the surroundings,and the trabsformation is an isothermal one.
As for the problem itself,with these having been made clear,u can assume for the points a) and b) that the transformation is adiabatica/isentropical and apply the formula for a general transformation of this kind.Don' forget that the adiabatical exponent "\gamma" is calculated for a diatomical molecule and it is found to be 7/5.
For the point c),thing are simpler,because the law of transforlation gets the simple form:pV=const.

Good luck!

 

[wais]

The main difference between filling a tank rapidly and filling it slowly is the rate at which the oxygen is compressed. When filling rapidly, the oxygen is compressed adiabatically, meaning it is compressed quickly and without any heat exchange with the surroundings. This results in an increase in temperature and pressure in the tank.

In contrast, filling slowly means that the oxygen is compressed isothermally, meaning the temperature remains constant throughout the process. In this case, the pressure increases gradually as the oxygen is slowly compressed into the tank.

To answer the questions posed in the problem, we need to use the ideal gas law equation, which states that PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature.

a) To determine how much oxygen fills the tank when filled rapidly, we can use the ideal gas law and the given information about the tank's volume and pressure. Rearranging the equation to solve for n (number of moles), we get n = PV/RT. Plugging in the values for pressure, volume, and temperature, we get n = (122 atm * 11.4 L) / (0.0821 L*atm/mol*K * 280 K) = 5.84 moles of oxygen.

b) To find the final temperature of the oxygen when the tank is filled rapidly, we can use the adiabatic compression equation, PV^gamma = constant, where gamma is the adiabatic index (for oxygen, gamma = 1.4). Rearranging the equation to solve for T (temperature), we get T = (P*V^gamma) / (R*n). Plugging in the values for pressure, volume, gas constant, and number of moles from part (a), we get T = (122 atm * (11.4 L)^1.4) / (0.0821 L*atm/mol*K * 5.84 mol) = 396 K. This is the final temperature of the oxygen at the instant the tank is filled.

c) To determine how much oxygen is required to fill the tank slowly and isothermally, we can use the ideal gas law again and the given information about the tank's volume and pressure. However, this time the temperature will remain constant at 280 K. Rearranging the equation to solve for n, we get