Question

If the atmospheric pressure in a tank is 23 atmospheres at an altitude of 1,000 feet, the air temperature in the tank is 700F, and the volume of the tank is 800 f3, determine the weight of the air in the tank.

Answer 1

W = 289.70 kg

Step-by-step explanation:

Given data:

Pressure in tank = 23 atm

Altitude 1000 ft

Air temperature in tank T = 700 F

Volume of tank = 800 ft^3 = 22.654 m^3

from ideal gas equation we have

PV =n RT

Therefore number of mole inside the tank is

`(1)/(n) = (RT)/(PV)`

`= \frac{8.206* 10^(-5)} 644.261}{23* 22.654}`

`= 1.02* 10^(-4)`

`n = 10^4 mole`

we know that 1 mole of air weight is 28.97 g

therefore, tank air weight is
`W = 10^4* 28.91 g = 289700 g`

W = 289.70 kg