The following table displays the vapor pressure of a noble gas at several different temperature. Use the date in the table Temperature (°C) Pressure (torr) 389.0 673.2 -20.5 The following -5.00 789.7 2.00 25.0 1166.8 55.0 1890.0
The Correct Answer and Explanation is :
The table provided lists the vapor pressures of a noble gas at various temperatures:
| Temperature (°C) | Pressure (torr) |
|---|---|
| -20.5 | 389.0 |
| -5.00 | 673.2 |
| 2.00 | 789.7 |
| 25.0 | 1166.8 |
| 55.0 | 1890.0 |
To determine the boiling point of this noble gas at 1.10 atm, we can use the Clausius-Clapeyron equation, which relates the change in vapor pressure with temperature to the heat of vaporization:
[ \ln \left( \frac{P_2}{P_1} \right) = -\frac{\Delta H_{\text{vap}}}{R} \left( \frac{1}{T_2} – \frac{1}{T_1} \right) ]
Where:
- ( P_1 ) and ( P_2 ) are the initial and final pressures, respectively.
- ( T_1 ) and ( T_2 ) are the initial and final temperatures (in Kelvin).
- ( \Delta H_{\text{vap}} ) is the heat of vaporization.
- ( R ) is the universal gas constant (8.314 J/mol·K).
Given that 1.10 atm is approximately 831.6 torr, we can use the data points closest to this pressure to estimate the boiling point.
From the table, the pressures at 25.0°C and 55.0°C are 1166.8 torr and 1890.0 torr, respectively.
Using these values, we can calculate the heat of vaporization (( \Delta H_{\text{vap}} )) and then determine the boiling point at 1.10 atm.
However, without the exact heat of vaporization value, we can estimate the boiling point by interpolation.
The boiling point at 1.10 atm will be between 25.0°C and 55.0°C, closer to 25.0°C due to the pressure values.
Therefore, the boiling point of this noble gas at 1.10 atm is approximately 25.0°C.
This estimation assumes that the heat of vaporization remains relatively constant over the temperature range considered.
For more precise calculations, the exact heat of vaporization and additional data points would be necessary.