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3 years ago

14

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, a

nd the volume of the tank is 800 f3, determine the weight of the air in the tank.

1 answer:

liraira [26]3 years ago

6 0

Answer:

W = 289.70 kg

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

$\frac{1}{n} = \frac{RT}{PV}$

$= \frac{8.206\times 10^{-5}} 644.261}{23\times 22.654}$

$= 1.02\times 10^{-4}$

$n = 10^4 mole$

we know that 1 mole of air weight is 28.97 g

therefore, tank air weight is $W = 10^4\times 28.91 g = 289700 g$

W = 289.70 kg

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