Question

Calculate the new temperature, in degrees Celsius, for each of the following with \(n\) and \(V\) constant: a. A sample of xenon at \(25^{\circ} \mathrm{C}\) and \(740 \mathrm{mmHg}\) is cooled to give a pressure of \(620 \mathrm{mmHg}\). b. A tank of argon gas with a pressure of \(0.950\) atm at \(-18{ }^{\circ} \mathrm{C}\) is heated to give a pressure of 1250 torr.

Short answer

a.\( -23.15^{\text{o}} \text{C} \) , b. \( 168.71^{\text{o}} \text{C} \)

Step-by-step solution

- Understanding the initial conditions and the formula

Use the ideal gas law in the form of combined gas law for processes where the temperature, pressure, and volume change: \[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \] where \(P\) is the pressure, \(T\) is the temperature in Kelvin, and the subscripts 1 and 2 refer to the initial and final states, respectively.

- Convert the temperature to Kelvin (Part a)

The initial temperature is given as \(25^{\text{o}} \text{C}\). Convert it to Kelvin. \[ T_1 = 25^{\text{o}} \text{C} + 273.15 = 298.15 \text{ K} \]

- Identify initial and final pressures (Part a)

The initial pressure \( P_1 \) is \( 740 \text{ mmHg} \) and the final pressure \( P_2 \) is \( 620 \text{ mmHg} \).

- Apply the combined gas law (Part a)

Rearrange the formula to solve for \( T_2 \): \[ T_2 = T_1 \times \frac{P_2}{P_1} \] Substitute the known values: \[ T_2 = 298.15 \text{ K} \times \frac{620 \text{ mmHg}}{740 \text{ mmHg}} \] Calculate the final temperature: \[ T_2 = 250 \text{ K} \]

- Convert the temperature back to Celsius (Part a)

Convert the final temperature back to Celsius: \[ T_2 = 250 \text{ K} - 273.15 = -23.15^{\text{o}} \text{C} \]

- Convert the temperature to Kelvin (Part b)

The initial temperature is given as \(-18^{\text{o}} \text{C}\). Convert it to Kelvin. \[ T_1 = -18^{\text{o}} \text{C} + 273.15 = 255.15 \text{ K} \]

- Identify initial and final pressures (Part b)

The initial pressure \( P_1 \) is\( 0.950 \text{ atm} \) and the final pressure \( P_2 \) is \( 1250 \text{ torr} \). Convert the pressures to the same unit (atm or torr). 1 atm = 760 torr. \[ P_1 = 0.950 \text{ atm} \] \[ P_2 = 1250 \text{ torr} \times \frac{1 \text{ atm}}{760 \text{ torr}} = 1.645 \text{ atm} \]

- Apply the combined gas law (Part b)

Rearrange the formula to solve for \( T_2 \): \[ T_2 = T_1 \times \frac{P_2}{P_1} \] Substitute the known values: \[ T_2 = 255.15 \text{ K} \times \frac{1.645 \text{ atm}}{0.950 \text{ atm}} \] Calculate the final temperature: \[ T_2 = 441.86 \text{ K} \]

- Convert the temperature back to Celsius (Part b)

Convert the final temperature back to Celsius: \[ T_2 = 441.86 \text{ K} - 273.15 = 168.71^{\text{o}} \text{C} \]

Key concepts

ideal gas law

In order to understand the combined gas law, first, we need to grasp the basics of the ideal gas law. The ideal gas law is represented by the equation: \[ PV = nRT \] Here,

• \(P\) = Pressure
• \(V\) = Volume
• \(n\) = Number of moles of gas
• \(R\) = Ideal gas constant
• \(T\) = Temperature in Kelvin

This law helps us understand how gases behave under different conditions. The combined gas law is derived from the ideal gas law and relates the initial and final states of a gas sample when the number of moles remains constant. In the combined gas law, we use: \[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \] This equation allows us to solve problems where the pressure and temperature of a gas change, while its volume and amount remain constant.

temperature conversion

Temperature conversion is a crucial step in solving gas law problems. The combined gas law requires the temperature to be in Kelvin. The conversion formulas are:

• °C to Kelvin: \( T(K) = T(°C) + 273.15 \)
• Kelvin to °C: \( T(°C) = T(K) - 273.15 \)

For example, if we have a temperature of \( 25^{\circ} \text{C} \), it needs to be converted to Kelvin: \[ 25^{\circ} \text{C} + 273.15 = 298.15 \text{ K} \] This step is vital since calculations involving gas laws must use absolute temperature values to be accurate.

pressure units conversion

Different pressure units can be used in problems, and converting between them is key. Common units include atmospheres (atm), millimeters of mercury (mmHg), and torr.

• 1 atm = 760 mmHg
• 1 atm = 760 torr

If we need to convert pressure values for consistent units in equations, we use these conversions. For instance, to convert 1250 torr to atm: \[ 1250 \text{ torr} \times \frac{1 \text{ atm}}{760 \text{ torr}} = 1.645 \text{ atm} \] Using the same units helps avoid mistakes in calculations and ensures correct results.

Kelvin to Celsius

Once we have solved for the temperature in Kelvin using the combined gas law, the final step is often to convert this back to degrees Celsius, especially when the problem specifically requests it. The conversion is simple: \[ T(^{\circ} \text{C}) = T(K) - 273.15 \] For example, if the final temperature calculated is \( 441.86 \text{ K} \), convert it to degrees Celsius as: \[ 441.86 \text{ K} - 273.15 = 168.71^{\circ} \text{C} \] This conversion is important for providing the answer in a standard and understandable temperature scale, often preferred in real-world scenarios.