# A rigid tank contains 1.40 moles of nitrogen, which can be treated as an ideal gas, at a pressure of 22.6 atm. ?

A rigid tank contains 1.40 moles of nitrogen, which can be treated as an ideal gas, at a pressure of 22.6 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 4.30 atm. How many moles are removed?

What If? In a separate experiment beginning from the same initial conditions, including a temperature Ti of 25.0°C, half the number of moles found in part (a) are withdrawn while the temperature is allowed to vary and the pressure undergoes the same change from 22.6 atm to 4.30 atm. What is the final temperature (in °C) of the gas?

### 2 Answers

- ?Lv 72 months agoFavourite answer
(a) The remaining moles are (4.30)*(1.40 mol)/(22.6) = 0.2664 moles, so the moles removed were about 1.13 moles.

(b) In this one, you can figure out the volume of the tank from pV = nRT and use the initial conditions (with T = 298.15K). Then put n = 0.833 moles, p = 4.30 atm, use the volume that you just figured out; and solve for T.

- Anonymous2 months ago
Assume that y mol of N₂ are removed.

Initial: P₁ = 22.6 atm, n₁ = 1.40 mol

Final: P₂ = 4.30 atm, n₂ = (1.40 - n) mol

PV = nRT

When V and T are constant: P/n = RT/V = constant

Hence, P₁/n₁ = P₂/n₂

Then, n₂ = n₁ × (P₂/P₁)

1.40 - y = 1.40 × (4.30/22.6)

1.40 - y = 0.266

y = 1.134 (to 3 sig. fig.)

Hence, 1.134 mol N₂ is removed.

====

Initial: P₁ = 22.6 atm, n₁ = 1.40 mol, T₁ = (273 + 25) K = 298 K

Final: P₂ = 4.30 atm, n₂ = [1.40 - (1.134/2)] mol = 0.833 mol, T₂ = ? K

PV = nRT

When V is constant: nT/P = V/R = constant

Hence, n₁T₁/P₁ = n₂T₂/P₂

Then, T₂ = T₁ × (P₂/P₁) × (n₁/n₂)

Final temperature, T₂ = 298 × (4.30/22.6) × (1.40/0.833) K = 95.3 K = -177.7°C