- Convert temperature: \(T = 0.00 + 273.15 = 273.15 \, \text{K}\).
- Rearrange the Ideal Gas Law: \(V = \frac{nRT}{P}\).
- Calculate the volume: \(V = \frac{0.500 \times 8.31 \times 273.15}{65.0} \approx 17.46 \, \text{L}\).
- Final Answer: \(\boxed{17.5}\) L.
Explanation
- Analysis and Data Preparation
To find the original volume of the Argon gas, we start with the ideal gas law:
\[PV = nRT\]
where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is the absolute temperature in Kelvin.
We are given the following information:
- Amount of substance \(n = 0.500 \, \text{mol}\)
- Pressure \(P = 65.0 \, \text{kPa}\)
- Temperature \(T = 0.00^{\circ}\text{C}\)
- Gas constant \(R = 8.31 \, \text{L} \cdot \text{kPa} / (\text{mol} \cdot \text{K})\)
First, we must convert the temperature from Celsius to Kelvin:
\[T(\text{K}) = T(^{\circ}\text{C}) + 273.15\]
\[T = 0.00 + 273.15 = 273.15 \, \text{K}\]
- Calculation of Volume
Now that we have all the values in appropriate units, we rearrange the ideal gas law to solve for volume \(V\):
\[V = \frac{nRT}{P}\]
Substituting the known values into the equation:
\[V = \frac{0.500 \, \text{mol} \times 8.31 \, \text{L} \cdot \text{kPa} / (\text{mol} \cdot \text{K}) \times 273.15 \, \text{K}}{65.0 \, \text{kPa}}\]
Calculating the numerator:
\[0.500 \times 8.31 \times 273.15 = 1134.93825\]
Dividing by the pressure \(65.0 \, \text{kPa}\):
\[V = \frac{1134.93825}{65.0} \approx 17.4606 \, \text{L}\]
Rounding to three significant figures, we get \(V = 17.5 \, \text{L}\).
Examples
The ideal gas law is essential in many industrial and scientific fields. For example, engineers designing airbag deployment systems in vehicles must calculate the precise amount of gas needed to fill the bag at a specific pressure and temperature to ensure passenger safety. By understanding the relationship between pressure, volume, and temperature, they can reliably determine how much gas is required for optimal performance, ensuring the bag inflates correctly within milliseconds.