A container is filled with an ideal gas to a pressure of 11.0 atm at \(0^{\circ} \mathrm{C} .\) a. What will be the pressure in the container if it is heated to $45^{\circ} \mathrm{C} ?$ b. At what temperature would the pressure be 6.50 atm? c. At what temperature would the pressure be 25.0 atm?

First, we need to convert Celsius temperatures to Kelvin for all given temperatures. Initial temperature, \(T_1\): \(T_1 = 0^{\circ} C + 273.15 = 273.15 K\) Temperature for part (a), \(T_2\): \(T_2 = 45^{\circ} C + 273.15 = 318.15 K\)

a. Calculate the final pressure, \(P_2\), when temperature is increased to \(45^{\circ} C\): We can rearrange the formula to solve for the final pressure, this gives us: \(P_2 = P_1 \times \frac{T_2}{T_1}\) Substitute known values: \(P_2 = 11.0 \, atm \times \frac{318.15 K}{273.15 K} \) Now, calculate the pressure: \(P_2 ≈ 12.76\, atm\) b. What temperature would the pressure be 6.50 atm? Rearrange the formula to solve for temperature as : \(T_2 = \frac{T_1 \times P_2}{P_1}\) Plugging in known values: \(T2 = \frac{273.15 K \times 6.50 \, atm}{11.0\, atm}\) Now, calculate the temperature and return it to Celsius: \(T_2 ≈ 128.54 K \) \(T_2^{\circ}C =128.54 K - 273.15 = -144.61^{\circ}C \) c. At what temperature would the pressure be 25.0 atm? For part (c), use the same rearranged formula: \(T_2 = \frac{T_1 \times P_2}{P_1}\) Replacing the known values: \(T_2 = \frac{273.15 K \times 25.0\,atm}{11.0\, atm}\) Now, calculate the temperature and express it in Celsius: \(T_2 ≈ \) 621.64 K \(T_2^{\circ}C = 621.64 K - 273.15 = 348.49^{\circ}C \) So the solutions are: a. Pressure at \(45^{\circ}C\) will be about 12.76 atm; b. The temperature for a pressure of 6.50 atm would be approximately \(-144.61^{\circ}C\); c. The temperature for a pressure of 25.0 atm would be approximately \(348.49^{\circ}C\).