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Bezzdna [24]
3 years ago
6

Heliox is a helium‑oxygen mixture that may be used in scuba tanks for divers working at great depths. It is also used medically

as a breathing treatment. A 7.25 L tank holds helium gas at a pressure of 1906 psi. A second 7.25 L tank holds oxygen at a pressure of 506.7 psi. The two gases are mixed in a 7.25 L tank. If the temperature remains the same throughout the process, what is the pressure of the gas mixture in the tank? Assume ideal gas behavior.
Physics
1 answer:
maksim [4K]3 years ago
7 0

Answer:

2412 psi is the pressure of the gas mixture in the tank.

Explanation:

Pressure of the helium gas = 1,906 psi

Volume of the helium gas = 7.25 L

Pressure of the oxygen gas = 506 psi

Volume of the oxygen gas = 7.25 L

After mixing both gases in a container with volume 7.25 L at with constant temperature.

Since, the temperature and volume remained constant, pressure becomes directly dependent on moles of gases.So, when we mix gases together the moles of gases will also add and along with that pressure of individual gas will also get added to give total pressure of the mixture in a tank.

Total pressure = pressure(Heluim)+pressure(oxygen)

P_{total}=1,906 psi+506 psi=2,412 psi

2412 psi is the pressure of the gas mixture in the tank.

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Answer:

a) v₁fin = 3.7059 m/s   (→)

b) v₂fin = 1.0588 m/s     (→)

Explanation:

a) Given

m₁ = 0.5 Kg

L = 70 cm = 0.7 m

v₁in = 0 m/s   ⇒  Kin = 0 J

v₁fin = ?

h<em>in </em>= L = 0.7 m

h<em>fin </em>= 0 m   ⇒    U<em>fin</em> = 0 J

The speed of the ball before the collision can be obtained as follows

Einitial = Efinal

⇒ Kin + Uin = Kfin + Ufin

⇒ 0 + m*g*h<em>in</em> = 0.5*m*v₁fin² + 0

⇒ v₁fin = √(2*g*h<em>in</em>) = √(2*(9.81 m/s²)*(0.70 m))

⇒ v₁fin = 3.7059 m/s   (→)

b)  Given

m₁ = 0.5 Kg

m₂ = 3.0 Kg

v₁ = 3.7059 m/s    (→)

v₂ = 0 m/s

v₂fin = ?

The speed of the block just after the collision can be obtained using the equation

v₂fin = 2*m₁*v₁ / (m₁ + m₂)

⇒  v₂fin = (2*0.5 Kg*3.7059 m/s) / (0.5 Kg + 3.0 Kg)

⇒  v₂fin = 1.0588 m/s     (→)

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