Question

1. How many moles of oxygen will occupy a volume of 2.5 liters at 1.2 atm and 25 °C?

Answer 1

PV = V (1.2 n. 0821) (298) = 123 mols 2.

Step-by-step explanation:

We need to use the ideal gas law for this problem, which states that:

PV=nRT

P is pressure in atmospheric pressures (in this case)

V is volume in liters (in this case)

n is moles of the substance

R is the gas constant (for this case, we use 0.082l atm mol−1K−1)

T is the temperature in Kelvin (as always)

Plugging in the known values, V=2.5l, P=1.2atm, T=298K, we get

n=PVRT=1.2⋅2.5298⋅0.082≈0.122 moles

So, there will be 0.122 moles of oxygen gas.

Answer 2

The number of moles of oxygen in a 2.5-liter volume at 1.2 atm and 25 °C is approximately 0.122 moles, calculated using the ideal gas law equation.

To calculate the number of moles of oxygen, we can use the ideal gas law equation:

`\[PV = nRT\]`

Where:

`\(P\)` = pressure (in atm)

`\(V\)` = volume (in liters)

`\(n\)` = number of moles

`\(R\)` = ideal gas constant
`(\(0.0821 \: \text{atm} \cdot \text{L/mol} \cdot \text{K}\))`

`\(T\)` = temperature (in Kelvin)

First, we need to convert the temperature to Kelvin using the formula T(K) = T(°C) + 273.15:

T = 25 °C + 273.15 = 298.15K

Now, plug the values into the ideal gas law:

`\[ (1.2 \: \text{atm}) \cdot (2.5 \: \text{L}) = n \cdot (0.0821 \: \text{atm} \cdot \text{L/mol} \cdot \text{K}) \cdot (298.15 \: \text{K})\]`

Solving for
`\(n\)`, the number of moles:

`\[ n = \frac{(1.2 \: \text{atm} \cdot 2.5 \: \text{L})}{(0.0821 \: \text{atm} \cdot \text{L/mol} \cdot \text{K} \cdot 298.15 \: \text{K})} \approx 0.122 \: \text{moles}\]`

So, 0.122 moles of oxygen will occupy a volume of 2.5 liters at 1.2 atm and 25 °C.