Question

Two tanks (tank A and tank B) of gas are connected by a closed valve. Tank A is 5 liters and contains O2 gas at a pressure of 24 atm. Tank B is 3 liters and contains N2 gas at a pressure of 32 atm. Both tanks are held at the same temperature. The valve between the two tanks is opened and the gases are allowed to mix. After the gases have had time to mix, what is the partial pressure of the oxygen gas?

Answer 1

Answer:

PO2 = 6.7326 KPa....partial pressure

Explanation:

tank A (O2):

∴ V = 5 L

∴ P = 24 atm

assuming: T = 298 K constant and ideal gas

⇒ nO2 = ((24 atm)(5 L))/((0.082 atm.L/K.mol)(298 K) = 4.91 mol O2

Tank B (N2):

∴ V = 3 L

∴ P = 32 atm

⇒ nN2 = ((32 atm)(3L))/((0.082 atm.L/K.mol)(298 K)) = 3.928 mol

moles mix:

⇒ n = nO2 + nN2 = 4.91 mol O2 + 3.928 mol N2 = 8.8386 mol mix

raoult's law:

• PO2 = (XO2)(P*O2)....partial pressure of the O2

∴ XO2 = nO2/nmix = 4.91/8.8386 = 0.555......molar fraction O2

∴ P*O2 .....vapor pressure of the pure component at T(K):

⇒ Antoine:

• LnP(KPa) = A - B/(T(K)+C)....Antoine ecuation

for the O2:

∴ A = 5.15038;  B = 780.259;  C = - 4.0758

⇒ Ln P*O2 = 5.15038 - ( 780.259 / ( 298 - 4.0758))

⇒ Ln P*O2 = 2.496

⇒ P*O2 = 12.131 KPa

⇒ PO2 = (0.555)(12.131 KPa) = 6.7326 KPa

Answer 2

Answer:

15 atm

Explanation:

Tank A is V₁ = 5 liters and contains O₂ gas at a pressure (P₁) of 24 atm. When it is connected to tank B, the total volume V₂ is 5 L + 3 L = 8 L. We can calculate the final pressure (P₂) of O₂ using Boyle's law (assuming ideal behavior and constant temperature).

$P_{1}.V_{1}=P_{2}.V_{2}\\P_{2}=\frac{P_{1}.V_{1}}{V_{2}} =\frac{24atm.5L}{8L} =15atm$

The pressure of O₂ is independent of the pressure of N₂.