Thermodynamic Question: Calculating Air Mass & Heat Transfer in a 2m3 Tank

[SPDupuis03]

Please help...


A stationary, un-insulated 2m3 rigid tank initially contains air at 100 kPa and 22oC. Air is available in an attached supply line at constant values of 600 kPa and 22oC. The valve to the supply line is opened and air enters the tank until the tank pressure reaches the supply line pressure. At that point, the valve is closed and the final temperature in the tank is 77oC. Determine the mass of air that entered the tank and the amount of heat transfer. Hint: u = Cv T, h = CpT

 

[Valkarie]

Mass of Air = 2 m3 x 100 kPa / 600 kPa x 0.02897 kg/m3 = 0.009616 kg Heat Transfer = mass x specific heat capacity x change in temperature = 0.009616 kg x 1.005 kJ/kgK x (77oC - 22oC) = 4.36 kJ

 

[piercas]

+ Pv

To calculate the mass of air that entered the tank, we can use the ideal gas law, PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature. Rearranging for n, we get n = PV/RT.

We know the initial pressure and temperature in the tank, as well as the final pressure and temperature. We can use these values to calculate the number of moles of air that entered the tank.

n1 = (100 kPa)(2 m3) / (8.314 J/mol*K)(295 K) = 0.0081 moles

n2 = (600 kPa)(2 m3) / (8.314 J/mol*K)(350 K) = 0.0195 moles

The difference between these two values is the number of moles that entered the tank:

n = n2 - n1 = 0.0114 moles

To convert this to mass, we can use the molar mass of air, which is approximately 28.97 g/mol.

Mass = n * M = (0.0114 moles)(28.97 g/mol) = 0.33 g

Therefore, approximately 0.33 grams of air entered the tank.

To calculate the amount of heat transfer, we can use the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system.

ΔU = Q - W

Since the tank is stationary, there is no work done, so W = 0. Therefore,

ΔU = Q

We can calculate the change in internal energy using the specific heat capacity at constant volume (Cv) and the change in temperature (ΔT).

ΔU = mCvΔT

We know the final temperature in the tank and the initial temperature, so we can calculate the change in temperature as:

ΔT = T2 - T1 = (350 K) - (295 K) = 55 K

We also know that Cv = R/γ, where R is the gas constant and γ is the specific heat ratio. For air, γ = 1.4.

Cv = 8.314 J/mol*K / 1.