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.

we can conclude that the melting point of table salt is above room temperature (it's actually 801 deg c). the melting point of mercury must be below room temperature (- 39 deg c).