Question

A metal tank containing 7.75 moles of oxygen is at 295 K with an internal pressure of 179

atmospheres. What is the volume of this tank at these conditions?
I NEED HELP ASAP

Answer 1

The volume of the tank can be calculated using the ideal gas law, which states that PV = nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the ideal gas constant, and T is temperature.

In this case, we have P = 179 atm, n = 7.75 moles, R = 0.08206 Latm/(Kmol), and T = 295 K. Plugging these values into the ideal gas law equation and solving for V gives us:

V = (nRT)/P V = (7.75 moles * 0.08206 Latm/(Kmol) * 295 K) / (179 atm) V ≈ 1.01 L

So the volume of this tank at these conditions is approximately 1.01 liters.

Step-by-step explanation:

Answer 2

Volume = 1.05 L (3 s.f.)

Step-by-step explanation:

To find the volume of the tank, we can use the ideal gas law.

Ideal Gas Law

`\boxed{\sf PV=nRT}`

where:

• P is the pressure measured in atmospheres (atm).
• V is the volume measured in liters (L).
• n is the number of moles.
• R is the ideal gas constant (0.082057366080960 L atm mol⁻¹ K⁻¹).
• T is the temperature measured in kelvin (K).

The given values are:

• P = 179 atm
• n = 7.75 mol
• R = 0.082057366080960 L atm mol⁻¹ K⁻¹
• T = 295 K

Substitute the given values into the formula and solve for V:

`\implies \sf 179 \cdot V=7.75 \cdot 0.082057366080960\cdot 295`

`\implies \sf V=(7.75 \cdot 0.082057366080960\cdot 295)/(179)`

`\implies \sf V=(187.6036532 \dots )/(179)`

`\implies \sf V=1.04806510 \dots\;L`

`\implies \sf V=1.05\;L\;(3\;s.f.)`

Therefore, the volume of the tank in these conditions is 1.05 liters (3 s.f.).