To calculate the pressure of the oxygen gas, we will use the Ideal Gas Law:

$PV = nRT$

Where:

• $P$ is the pressure in atm (which we will convert to mmHg later),
• $V$ is the volume in liters (50.0 L),
• $n$ is the number of moles of oxygen gas,
• $R$ is the ideal gas constant (0.0821 L·atm/mol·K),
• $T$ is the temperature in Kelvin.

Step 1: Convert mass of oxygen gas to moles

The molar mass of $O_2$ (oxygen gas) is $32.00 \, \text{g/mol}$.

$n = \frac{\text{mass}}{\text{molar mass}} = \frac{20.0 \, \text{g}}{32.00 \, \text{g/mol}} = 0.625 \, \text{mol}$

Step 2: Convert temperature to Kelvin

$T = 25^\circ C + 273.15 = 298.15 \, \text{K}$

Step 3: Apply the Ideal Gas Law

Now, substitute the values into the Ideal Gas Law:

$P \times 50.0 = 0.625 \times 0.0821 \times 298.15$

Solving for $P$:

$P = \frac{0.625 \times 0.0821 \times 298.15}{50.0}$

Step 4: Convert pressure from atm to mmHg

1 atm = 760 mmHg, so we'll multiply the pressure in atm by 760 to convert to mmHg.

Let me calculate that.The pressure of the oxygen gas in the 50.0 L container at 25°C is approximately 232.54 mmHg.

Would you like further details or have any questions?

Here are 5 related questions to explore:

1. How would the pressure change if the container's volume were halved?
2. What happens to the pressure if the temperature increases to 50°C?
3. How would the pressure change if you doubled the amount of oxygen?
4. What if the gas was nitrogen instead of oxygen? Would the pressure be different?
5. How does the Ideal Gas Law differ from the Real Gas Law under high pressures?

Tip: Always convert temperatures to Kelvin when using gas laws like the Ideal Gas Law.