Question

A 2.30 L container holds 24.00 grams of oxygen gas at 108.5 kPa at 22.00oC. What will be the final volume of the container if the pressure increases by 4.20 kPa, the amount increases to 1.85 moles, and the temperature increases to 52.00oC?

Answer 1

To find the final volume of the container after the changes in pressure, moles, and temperature, we can use the Ideal Gas Law which is given by the equation:

\[
PV = nRT
\]

where:
- \( P \) is the pressure in kPa,
- \( V \) is the volume in liters,
- \( n \) is the number of moles,
- \( R \) is the ideal gas constant (8.314 J/(mol·K) or 8.314 kPa·L/(mol·K)),
- \( T \) is the temperature in Kelvin.

### Step 1: Convert all necessary values to appropriate units

1. **Initial Pressure (P1)**:
\[
P_1 = 108.5 \text{ kPa} + 4.20 \text{ kPa} = 112.7 \text{ kPa}
\]

2. **Final Moles (n)**:
\[
n = 1.85 \text{ moles}
\]

3. **Final Temperature (T)**:
\[
T = 52.00^\circ C = 52.00 + 273.15 = 325.15 \text{ K}
\]

### Step 2: Use the Ideal Gas Law to find the final volume (V)

Rearranging the Ideal Gas Law to solve for volume:

\[
V = \frac{nRT}{P}
\]

### Step 3: Calculate the final volume

Using \( R = 8.314 \text{ kPa} \cdot \text{L}/(\text{mol} \cdot \text{K}) \):

\[
V = \frac{(1.85 \text{ mol}) \times (8.314 \text{ kPa} \cdot \text{L}/(\text{mol} \cdot \text{K})) \times (325.15 \text{ K})}{112.7 \text{ kPa}}
\]

### Step 4: Perform the calculation

Calculating the numerator first:

\[
1.85 \text{ mol} \times 8.314 \text{ kPa} \cdot \text{L}/(\text{mol} \cdot \text{K}) \times 325.15 \text{ K} \approx 4,785.56045 \text{ kPa} \cdot \text{L}
\]

Now, calculating the final volume:

\[
V = \frac{4,785.56045 \text{ kPa} \cdot \text{L}}{112.7 \text{ kPa}} \approx 42.463 \text{ L}
\]

So, the final volume of the container is approximately:

\[
\boxed{42.46 \text{ L}}
\]