Question

What is the pressure in a 2.8 L container that has 0.652 moles of oxygen gas at 25.0 °C?

Answer 1

To find the pressure in the container, use the ideal gas law equation (PV = nRT). Plug in the given values and solve for P. The pressure in the 2.8 L container with 0.652 moles of oxygen gas at 25.0 °C is approximately 38.16 atm.

Step-by-step explanation:

To find the pressure in the container, we can use the ideal gas law equation, which is expressed as:

PV = nRT

Where:

• P is the pressure
• V is the volume
• n is the number of moles
• R is the ideal gas constant
• T is the temperature in Kelvin

In this case, we are given the volume of the container (2.8 L), the number of moles of oxygen gas (0.652 moles), and the temperature (25.0 °C). We need to convert the temperature to Kelvin by adding 273.15, so it becomes 298.15 K. The ideal gas constant (R) is 0.0821 L·atm/(mol·K).

Plugging in the values into the ideal gas law equation and solving for P:

P(2.8) = (0.652)(0.0821)(298.15)

P = (0.652)(0.0821)(298.15)/2.8

P ≈ 38.16 atm

Therefore, the pressure in the 2.8 L container with 0.652 moles of oxygen gas at 25.0 °C is approximately 38.16 atm.

Answer 2

To calculate the pressure of a gas using the ideal gas law, you can use the formula:

\[ PV = nRT \]

where:

- \( P \) is the pressure,

- \( V \) is the volume,

- \( n \) is the number of moles,

- \( R \) is the ideal gas constant (\(0.0821 \, \text{L} \cdot \text{atm} / \text{mol} \cdot \text{K}\)),

- \( T \) is the temperature in Kelvin.

First, convert the temperature from Celsius to Kelvin:

\[ T (\text{Kelvin}) = 25.0 + 273.15 = 298.15 \, \text{K} \]

Now, substitute the given values into the ideal gas law:

\[ P \times 2.8 \, \text{L} = (0.652 \, \text{mol}) \times (0.0821 \, \text{L} \cdot \text{atm} / \text{mol} \cdot \text{K}) \times (298.15 \, \text{K}) \]

Solve for \( P \):

\[ P = \frac{(0.652 \, \text{mol}) \times (0.0821 \, \text{L} \cdot \text{atm} / \text{mol} \cdot \text{K}) \times (298.15 \, \text{K})}{2.8 \, \text{L}} \]

\[ P \approx 4.91 \, \text{atm} \]

Therefore, the pressure in the 2.8 L container with 0.652 moles of oxygen gas at 25.0 °C is approximately 4.91 atm.

Step-by-step explanation: