- The Ideal Gas Law \(PV = nRT\) is used to calculate the gas pressure.
- Volume is converted from \(500.0 \text{ mL}\) to \(0.500 \text{ L}\) and temperature from \(265^\circ\text{C}\) to \(538.15 \text{ K}\).
- By rearranging the formula to \(P = \frac{nRT}{V}\) and substituting the values, we calculate the pressure.
- The final calculated pressure is \(\boxed{55.9 \text{ kPa}}\).
Explanation
- Analysis and Data Setup
We are given a gas sample with the following parameters:
- Moles of gas: \(n = 6.25 \times 10^{-3} \text{ mol}\)
- Volume of the flask: \(V = 500.0 \text{ mL}\)
- Temperature: \(T = 265^\circ\text{C}\)
- Ideal gas constant: \(R = 8.31 \text{ L} \cdot \text{kPa} / (\text{mol} \cdot \text{K})\)
Our objective is to calculate the pressure \(P\) of the gas in kilopascals (kPa) using the Ideal Gas Law formula: \(PV = nRT\). Before substituting the values into the formula, we must ensure all units are consistent with the ideal gas constant \(R\). The constant \(R\) uses Liters (L) and Kelvin (K), so we need to convert the volume from milliliters (mL) to Liters (L) and the temperature from Celsius (\$^\circ\text{C}\$) to Kelvin (K).
- Conversion and Calculation
First, convert the volume \(V\) from mL to L:
\[V = \frac{500.0 \text{ mL}}{1000} = 0.500 \text{ L}\]
Next, convert the temperature \(T\) from Celsius to Kelvin:
\[T = 265^\circ\text{C} + 273.15 = 538.15 \text{ K}\]
Now that the units are consistent, we can rearrange the Ideal Gas Law \(PV = nRT\) to solve for pressure \(P\):
\[P = \frac{nRT}{V}\]
Substitute the given values into the equation:
\[P = \frac{(6.25 \times 10^{-3} \text{ mol}) \times (8.31 \text{ L} \cdot \text{kPa} / (\text{mol} \cdot \text{K})) \times (538.15 \text{ K})}{0.500 \text{ L}}\]
Performing the calculation:
\[P = \frac{0.00625 \times 8.31 \times 538.15}{0.500} = \frac{27.950165625}{0.500} = 55.90033125 \text{ kPa}\]
Rounding to an appropriate number of significant figures (based on the input data), we get approximately \(55.9 \text{ kPa}\).
- Final Conclusion
The pressure of the gas, calculated using the Ideal Gas Law, is found to be \(\boxed{55.9 \text{ kPa}}\). The variables given in the problem were volume, temperature, and moles.
Examples
The Ideal Gas Law is a fundamental concept in chemistry and physics used in various practical scenarios. For instance, it is used by engineers to calculate the pressure inside a pressurized container (like a gas cylinder or a tire) when the temperature changes, ensuring that the container can withstand the internal stress. By understanding how the number of particles, volume, and temperature affect the pressure, one can ensure safety and optimize processes in industries ranging from aerospace engineering to scuba diving.