Question

Solve for the new pressure, in torr, for each of the following, if \(n\) and \(V\) are constant: a. A gas with an initial pressure of 1200 torr at \(155^{\circ} \mathrm{C}\) is cooled to \(0{ }^{\circ} \mathrm{C}\). b. A gas in an aerosol can with an initial pressure of \(1.40\) atm at \(12^{\circ} \mathrm{C}\) is heated to \(35^{\circ} \mathrm{C}\).

Short answer

a. 765 torr; b. 1148 torr

Step-by-step solution

- Understand the problem

This problem involves using the gas law, specifically the relationship between pressure and temperature when the number of moles (n) and the volume (V) of the gas are constant. The formula we will use is known as Gay-Lussac's Law: \[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \] where \( P_1 \) and \( T_1 \) are the initial pressure and temperature, and \( P_2 \) and \( T_2 \) are the final pressure and temperature. Note that the temperatures must be in Kelvin.

- Convert temperatures to Kelvin

Convert the given temperatures to Kelvin using the formula: \[ T(K) = T(^{\circ}C) + 273.15 \] For part a: \[ T_1 = 155^{\circ}C + 273.15 = 428.15 \: K \] \[ T_2 = 0^{\circ}C + 273.15 = 273.15 \: K \] For part b: \[ T_1 = 12^{\circ}C + 273.15 = 285.15 \: K \] \[ T_2 = 35^{\circ}C + 273.15 = 308.15 \: K \]

- Solve for the new pressure (Part a)

Using Gay-Lussac’s Law, rearrange to solve for the new pressure \( P_2 \): \[ P_2 = P_1 \cdot \frac{T_2}{T_1} \] Substitute the known values for part a: \[ P_2 = 1200 \: \mathrm{torr} \cdot \frac{273.15}{428.15} \approx 765 \: \mathrm{torr} \]

- Solve for the new pressure (Part b)

First, convert the initial pressure from atm to torr (1 atm = 760 torr): \[ P_1 = 1.40 \: \mathrm{atm} \cdot 760 \: \frac{\mathrm{torr}}{\mathrm{atm}} = 1064 \: \mathrm{torr} \] Using Gay-Lussac’s Law, substitute the known values for part b: \[ P_2 = 1064 \: \mathrm{torr} \cdot \frac{308.15}{285.15} \approx 1148 \: \mathrm{torr} \]

Key concepts

Pressure-Temperature Relationship

In the context of gases, the pressure-temperature relationship is a crucial concept. Gay-Lussac's Law states that for a fixed amount of gas at constant volume, the pressure of the gas is directly proportional to its absolute temperature. This means if you increase the temperature of a gas, its pressure increases if the volume remains unchanged, and vice versa. This relationship is expressed mathematically as: \[\frac{P_1}{T_1} = \frac{P_2}{T_2}\]. Here, \(P_1\) and \(T_1\) are the initial pressure and temperature, while \(P_2\) and \(T_2\) are the final pressure and temperature. Note, for the equation to work, temperatures must be in Kelvin. By understanding this concept, you can predict how changes in temperature will affect the pressure of a gas.

Ideal Gas Law

The Ideal Gas Law is a fundamental principle in understanding gases and their behavior. It combines several simple gas laws like Boyle's, Charles's, and Gay-Lussac's Laws. The Ideal Gas Law is written as: \[ PV = nRT \], where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the gas constant, and \(T\) is the temperature in Kelvin. This equation allows us to solve for any one of the variables if the other three are known. It's particularly helpful in many practical applications across chemistry and physics. In the provided problem, we specifically use Gay-Lussac's part of the Ideal Gas Law, focusing on the pressure-temperature relationship at constant volume and moles.

Temperature Conversion to Kelvin

In gas law calculations, it's essential to convert Celsius temperatures to Kelvin. The Kelvin scale starts at absolute zero, the point where all molecular motion stops. To convert Celsius to Kelvin, use the formula: \[ T(K) = T(^{\boxed{}}C) + 273.15 \]. For instance, to convert \(155^{\boxed{}}C\) to Kelvin, you add 273.15, resulting in 428.15 K. Similarly, \(0^{\boxed{}}C\) becomes 273.15 K. This conversion ensures you use the correct temperature scale that aligns with the principles of gas laws and avoids errors in your calculations, as both Gay-Lussac's Law and the Ideal Gas Law require temperatures in Kelvin.