- Calculate the initial temperature using the ideal gas law: \(T_1 = \frac{P_1 V}{n_1 R} \approx 122\text{ K}\).
- Calculate the final total moles of gas: \(n_2 = n_1 + 0.7 = 1.158\text{ mol}\).
- Calculate the final temperature using the ideal gas law: \(T_2 = \frac{P_2 V}{n_2 R} \approx 53.9\text{ K}\).
- Determine the change in temperature: \(\Delta T = T_2 - T_1 = \boxed{-68\text{ K}}\).
Explanation
- Problem Analysis and Setup
We are given a rigid bottle with an initial volume \(V = 2.50\text{ L}\). It contains \(n_1 = 0.458\text{ mol}\) of Helium gas at a pressure of \(P_1 = 1.83\text{ atm}\). After adding \(\Delta n = 0.7\text{ mol}\) of gas, the pressure increases to \(P_2 = 2.05\text{ atm}\). The total amount of gas is \(n_2 = n_1 + \Delta n = 0.458\text{ mol} + 0.7\text{ mol} = 1.158\text{ mol}\). We use the ideal gas law \(PV = nRT\) to find the initial and final temperatures, where \(R = 0.0821\text{ L}\cdot\text{atm}/(\text{mol}\cdot\text{K})\).
- Calculating Initial Temperature
The initial temperature \(T_1\) is calculated using \(T_1 = \frac{P_1 V}{n_1 R}\):
\[T_1 = \frac{1.83\text{ atm} \times 2.50\text{ L}}{0.458\text{ mol} \times 0.0821\text{ L}\cdot\text{atm}/(\text{mol}\cdot\text{K})} = \frac{4.575}{0.0376018} \approx 121.67\text{ K}\]
Rounding to three significant figures, we get \(T_1 = 122\text{ K}\).
- Calculating Final Temperature
The final temperature \(T_2\) is calculated using \(T_2 = \frac{P_2 V}{n_2 R}\):
\[T_2 = \frac{2.05\text{ atm} \times 2.50\text{ L}}{1.158\text{ mol} \times 0.0821\text{ L}\cdot\text{atm}/(\text{mol}\cdot\text{K})} = \frac{5.125}{0.0950718} \approx 53.91\text{ K}\]
Rounding to three significant figures, we get \(T_2 = 53.9\text{ K}\).
- Calculating the Change in Temperature
The change in temperature \(\Delta T\) is \(T_2 - T_1\):
\[\Delta T = 53.906\text{ K} - 121.670\text{ K} = -67.764\text{ K}\]
Applying the rules for significant figures in subtraction (using the decimal places of the values), the result should be rounded to the least precise decimal place among the temperatures, which is zero decimal places (from the initial temperature 122 K). Thus, \(\Delta T = -68\text{ K}\). The gas mixture cools down by 68 K.
Examples
Understanding how temperature and pressure change when adding gas to a container is vital in industrial processes such as chemical manufacturing and storage. For example, when filling a pressurized gas cylinder, engineers must calculate temperature changes to ensure that the container does not exceed its pressure rating. This application of the ideal gas law helps ensure safety and material integrity in various real-world scenarios.