Question

A 3.10 L tank holds 106.4 grams of fluorine gas at a pressure of 201 kPa. What was the original temperature in celsius if the container increases pressure to 275 kPa, the volume expands to 4.500 L and the amount increases to 5.50 moles?

Answer 1

To determine the original temperature of the fluorine gas in Celsius before the changes, we can use the ideal gas law:

\[
PV = nRT
\]

Where:
- \(P\) = pressure in pascals (Pa)
- \(V\) = volume in cubic meters (m³)
- \(n\) = number of moles of gas
- \(R\) = ideal gas constant, \(8.314 \, \text{J/(mol·K)}\)
- \(T\) = temperature in Kelvin (K)

Let's first convert everything to the correct units.

1. **Original conditions:**
- Volume: \(V = 3.10 \, \text{L} = 3.10 \times 10^{-3} \, \text{m}^3\)
- Pressure: \(P = 201 \, \text{kPa} = 201,000 \, \text{Pa}\)
- Mass of fluorine: \(m = 106.4 \, \text{g} = 0.1064 \, \text{kg}\)
- Molar mass of fluorine (F₂) = 19.00 g/mol × 2 = 38.00 g/mol
- Number of moles:
\[
n = \frac{m}{\text{molar mass}} = \frac{106.4 \, \text{g}}{38.00 \, \text{g/mol}} \approx 2.8 \, \text{moles}
\]

Using the ideal gas law to find the original temperature in Kelvin:

\[
T = \frac{PV}{nR}
\]

Substituting in the values:

\[
T = \frac{(201,000 \, \text{Pa})(3.10 \times 10^{-3} \, \text{m}^3)}{(2.8 \, \text{mol})(8.314 \, \text{J/(mol \cdot K)})}
\]

Calculating \(T\):

First, calculate the numerator:

\[
201,000 \, \text{Pa} \times 3.10 \times 10^{-3} \, \text{m}^3 = 623.1 \, \text{Pa}\cdot\text{m}^3
\]

Next, calculate the denominator:

\[
(2.8 \, \text{mol})(8.314 \, \text{J/(mol \cdot K)}) \approx 23.292 \, \text{J/K}
\]

Now we can find \(T\):

\[
T = \frac{623.1}{23.292} \approx 26.7 \, \text{K}
\]

**Convert the temperature from Kelvin to Celsius:**

\[
T(°C) = T(K) - 273.15 = 26.7 - 273.15 \approx -246.45 \, °C
\]

Therefore, the original temperature in Celsius was approximately:

\[
\boxed{-246.45} \, °C
\]