Question

3. What volume would be occupied by 3.5 mol of oxygen gas at a pressure

of 768 mm Hg and a temperature fo 25°C?
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Answer 1

The volume occupied by 3.5 mol of oxygen gas at a pressure of 768 mm Hg and a temperature of 25°C is approximately 84.19 liters, calculated using the Ideal Gas Law.

Step-by-step explanation:

To find the volume occupied by 3.5 mol of oxygen gas at a pressure of 768 mm Hg and a temperature of 25°C, we can use the Ideal Gas Law, which is PV = nRT.

First, we need to convert the given pressure from mm Hg to atm, since the gas constant R is typically given in units of L atm/(mol K).

1 atm = 760 mm Hg

So, P (in atm) = 768 mm Hg × (1 atm / 760 mm Hg) = 1.0105 atm

Next, convert the temperature from °C to Kelvin:

T (in K) = 25°C + 273.15 = 298.15 K

The gas constant R = 0.0821 L atm/(mol K).

Now, we can rearrange the ideal gas equation to solve for V:

V = (nRT) / P

V = (3.5 mol × 0.0821 L atm/(mol K) × 298.15 K) / 1.0105 atm

V = 85.1085 L / 1.0105 atm

V = 84.19 L

The volume occupied by 3.5 mol of oxygen gas under the given conditions is approximately 84.19 liters.

Answer 2

The oxygen will occupied 84.7 L.

Step-by-step explanation:

Ideal gases are a simplification of real gases that is done to study them more easily. It is considered to be formed by point particles, do not interact with each other and move randomly. It is also considered that the molecules of an ideal gas, in themselves, do not occupy any volume.

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:

P*V = n*R*T

where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas. The universal constant of ideal gases R has the same value for all gaseous substances.

In this case:

• P= 768 mmHg
• V= ?
• n= 3.5 mol
• R= 62.36
  `(mmHg*L)/(mol*K)`
• T= 25 C= 298 K (being 0 C= 273 K)

Replacing:

768 mmHg* V= 3.5 mol* 62.36
`(mmHg*L)/(mol*K)` *298 K

Solving:

`V=(3.5 mol*62.36(mmHg*L)/(mol*K)*298 K )/(768 mmHg)`

V= 84.7 L

The oxygen will occupied 84.7 L.