A rigid tank contains 1.80 moles of nitrogen, which can be treated as an ideal gas, at a pressure of 25.5 atm. While the tank and gas maintain a constant volume and temperature, a number of moles are removed from the tank, reducing the pressure to 5.70 atm. How many moles are removed

1.4 moles.

Explanation:

The following data were obtained from the question:

Initial mole (n1) = 1.8 moles

Initial pressure (P1) = 25.5 atm

Final pressure (P2) = 5.7 atm

Number of mole removed =?

Next, we shall obtain an expression relating pressure and number of mole together.

This is illustrated below:

From the ideal gas equation:

PV = nRT

Divide both side by V

P = nRT /V

Divide both side by n

P/n = RT/V

But volume (V) and temperature (T) are constant.

P/n = constant

P1/n1 = P2/n2

P1 is initial pressure.

P2 is final pressure.

n1 is the initial mole.

n2 is the final mole.

Next, we shall determine the final mole in tank as follow:

Initial mole (n1) = 1.8 moles

Initial pressure (P1) = 25.5 atm

Final pressure (P2) = 5.7 atm

Final mole (n2) =.?

P1/n1 = P2/n2

25.5/1.8 = 5.7/n2

Cross multiply

25.5 x n2 = 1.8 x 5.7

Divide both side by 25.5

n2 = (1.8 x 5.7) /25.5

n2 = 0.40 mole.

Therefore, the final mole in the tank is 0.40 mole.

Finally, we shall determine the number of mole that was removed as follow:

Initial mole (n1) = 1.8 moles

Final mole (n2) = 0.4 mole

Number of mole removed =?

The number of mole removed from the tank can be obtained by finding the difference between the initial mole and final mole as illustrated below:

Number of mole removed = initial mole – final mole

Number of mole removed = n1 – n2

Number of mole removed = 1.8 – 0.4

Number of mole removed = 1.4 moles

Therefore, 1.4 moles were removed from the tank.