Question

A liquid is placed in a 25.0-L flask. At \(140{ }^{\circ} \mathrm{C}\), the liquid evaporates completely to give a pressure of \(0.900 \mathrm{~atm}\). If the flask can withstand pressures up to \(1.30 \mathrm{~atm}\), calculate the maximum temperature to which the gas can be heated without breaking the flask.

Short answer

The maximum temperature is approximately 598 K or 325 °C.

Step-by-step solution

- Use the Ideal Gas Law

The ideal gas law is given by \[ PV = nRT \]. First, convert the temperature from Celsius to Kelvin: \[ T_1 = 140 + 273.15 = 413.15 \text{ K} \].

- Solve for the number of moles (n)

We know the pressure (\[ P_1 = 0.900 \text{ atm} \]), the volume (\[ V = 25.0 \text{ L}\]), and the temperature (\[ T_1 = 413.15 \text{ K} \]). Use the ideal gas law to find the number of moles: \[ n = \frac{P_1V}{RT_1} \]. Where \[ R = 0.0821 \frac{atm⋅L}{mol⋅K} \]. Substitute and solve for n: \[ n = \frac{0.900 \times 25.0}{0.0821 \times 413.15} \approx 0.664 \text{ moles} \].

- Set up the equation for the new conditions

We are given the maximum pressure (\[ P_2 = 1.30 \text{ atm} \]). Using the ideal gas law again, set up the equation with the new temperature (\[ T_2 \]) and solve: \[ P_2 V = n R T_2 \].

- Solve for the maximum temperature

Rearrange the equation to solve for \[ T_2 \]: \[ T_2 = \frac{P_2 V}{n R} \]. Substitute the known values: \[ T_2 = \frac{1.30 \times 25.0}{0.664 \times 0.0821} \approx 598 \text{ K} \]. Finally, convert back to Celsius if needed: \[ T_2 - 273.15 \approx 325 { }^{\text{°C}} \].

Key concepts

Pressure and Temperature Relationship

The relationship between pressure and temperature in a gas is crucial to understanding the behavior of gases.
According to Gay-Lussac's Law, for a given mass and constant volume, the pressure of a gas is directly proportional to its absolute temperature.

This implies that as the temperature of a gas increases, its pressure increases as well, provided the volume remains constant.
Mathematically, Gay-Lussac's Law can be expressed as:
\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \]
where \( P_1 \) and \( T_1 \) are the initial pressure and temperature, and \( P_2 \) and \( T_2 \) are the final pressure and temperature.

In the given problem, we see how an increase in temperature (if unchecked) could potentially exceed the maximum pressure threshold of the flask. Hence, calculating the maximum temperature that the gas can reach without breaking the flask is essential.

Gas Laws

The behavior of gases can be described by several fundamental laws such as Boyle's Law, Charles's Law, and the Ideal Gas Law.
Among these, the Ideal Gas Law combines all the simple gas laws into one comprehensive equation:
\[ PV = nRT \]
Here, \( P \) stands for pressure, \( V \) is volume, \( n \) is the number of moles of gas, \( R \) is the universal gas constant, and \( T \) is the absolute temperature in Kelvin.

The Ideal Gas Law provides a good approximation for the behavior of many gases under a variety of conditions, though it assumes no interactions between the gas molecules and that the volume of the gas molecules themselves is negligible.
In the example exercise, this law is used to calculate the number of moles of the gas when the liquid evaporates and to determine the maximum temperature the gas can reach without the flask breaking.

Molecular Calculations

When dealing with gases, it is often necessary to calculate the number of molecules or moles involved.
In the given problem, after converting the temperature to Kelvin, the Ideal Gas Law is used to find the number of moles \( n \).
This is achieved by rearranging the equation to solve for \( n \):
\[ n = \frac{P_1 V}{RT_1} \]
By substituting the given values: \[ n = \frac{0.900 \times 25.0}{0.0821 \times 413.15} \approx 0.664 \]
We find that there are approximately 0.664 moles of gas.

This number of moles is then kept constant to find the maximum temperature \( T_2 \) the gas can achieve without increasing the pressure beyond 1.30 atm.
Rearranging the Ideal Gas Law for the new conditions:
\[ T_2 = \frac{P_2 V}{n R} \]
and substituting the values again:
\[ T_2 = \frac{1.30 \times 25.0}{0.664 \times 0.0821} \approx 598 \]
Converting this back to Celsius gives approximately 325 °C.
These molecular calculations are essential for predicting gas behavior in real-world applications.