Question

Solve for the new pressure, in atm, for each of the following, if \(n\) and \(V\) are constant: a. A gas with an initial pressure of \(1.20\) atm at \(75^{\circ} \mathrm{C}\) is cooled to \(-32^{\circ} \mathrm{C}\). b. A sample of \(\mathrm{N}_{2}\) with an initial pressure of \(780 . \mathrm{mmHg}\) at \(-75^{\circ} \mathrm{C}\) is heated to \(28^{\circ} \mathrm{C}\).

Short answer

Part (a): 0.83 atm; Part (b): 1.56 atm.

Step-by-step solution

Concept: Ideal Gas Law

Begin by recalling the relationship between pressure and temperature for a fixed amount of gas at constant volume, described by the equation \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \).

Convert Temperatures to Kelvin

Convert the given temperatures from Celsius to Kelvin using the conversion formula: \( T(\text{K}) = T(\text{°C}) + 273.15 \). For part (a), convert 75 °C and -32 °C. For part (b), convert -75 °C and 28 °C.

Initial and Final Temperatures for Part (a)

For part (a), convert 75 °C to K: \( T_1 = 75 + 273.15 = 348.15 \) K. Convert -32 °C to K: \( T_2 = -32 + 273.15 = 241.15 \) K.

Initial and Final Temperatures for Part (b)

For part (b), convert -75 °C to K: \( T_1 = -75 + 273.15 = 198.15 \) K. Convert 28 °C to K: \( T_2 = 28 + 273.15 = 301.15 \) K.

Use Ideal Gas Law for Part (a)

Using the formula \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \) and solving for the final pressure \( P_2 \), we get: \( P_2 = P_1 \times \frac{T_2}{T_1} \). Substitute the values for part (a): \( P_2 = 1.20 \text{ atm} \times \frac{241.15 \text{ K}}{348.15 \text{ K}} \). Calculate \( P_2 \).

Use Ideal Gas Law for Part (b)

Similarly, for part (b): Convert the initial pressure from mmHg to atm if necessary (1 atm = 760 mmHg). Use the formula \( P_2 = P_1 \times \frac{T_2}{T_1} \). Substitute the values: \( P_2 = 780 \text{ mmHg} \times \frac{301.15 \text{ K}}{198.15 \text{ K}} \). Calculate \( P_2 \) in mmHg and convert to atm if needed.

Perform Calculations

For part (a): \( P_2 = 1.20 \text{ atm} \times \frac{241.15}{348.15} \approx 0.83 \text{ atm} \). For part (b): \( P_2 = 780 \text{ mmHg} \times \frac{301.15}{198.15} \approx 1185 \text{ mmHg} = \frac{1185 \text{ mmHg}}{760 \text{ mmHg/atm}} \approx 1.56 \text{ atm} \).

Key concepts

pressure-temperature relationship

One of the key concepts in understanding gas behavior is the pressure-temperature relationship. This relationship is crucial when working with gases under different conditions. For an ideal gas at constant volume, the pressure and temperature have a direct proportional relationship. This is expressed by the equation: \ \ \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \). \ \ This means that if the temperature of a gas increases, its pressure will also increase, as long as the volume remains unchanged. Conversely, if the temperature decreases, the pressure will decrease. When solving problems involving gases, it’s important to ensure that temperatures are in the same units — typically Kelvin (K) — to use this equation correctly and avoid errors.

This principle helps us predict how changes in temperature will affect the pressure of a gas sample. For instance, if we know the initial pressure and temperature, and either temperature changes, we can calculate the new pressure by rearranging the equation to: \ \ \( P_2 = P_1 \times \frac{T_2}{T_1} \). \

temperature conversion to Kelvin

When working with the Ideal Gas Law, always convert temperatures to Kelvin. The Kelvin scale is the SI unit for temperature and starts at absolute zero, meaning it's directly proportional to the energy of particles in a gas. Celsius (°C) and Fahrenheit (°F) are not suitable for these calculations because they are relative scales and can lead to incorrect results.

The conversion process is simple:
1. Take the temperature in Celsius.
2. Add 273.15 to convert it to Kelvin.

For example:
- An initial temperature of 75 °C is converted as follows: 75 + 273.15 = 348.15 K.
- A lower temperature of -32 °C becomes: -32 + 273.15 = 241.15 K.
- For temperatures in negatives, like -75 °C: -75 + 273.15 = 198.15 K.
- For positive temperatures, like 28 °C: 28 + 273.15 = 301.15 K.

ideal gas law calculations

The Ideal Gas Law equation, \(PV = nRT\), helps us understand how pressure (P), volume (V), and temperature (T) affect a gas. For a constant volume and amount of gas, the relationship simplifies and can be used to solve problems involving changing conditions, such as:
\ \ \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \). \ \ Here’s how you would use this in a problem: Assume you have a gas at an initial pressure \( P_1 \) and temperature \( T_1 \). You want to find the new pressure \( P_2 \) after the temperature changes to \( T_2 \). The steps are:
1. Convert all temperatures to Kelvin.
2. Use the equation \( P_2 = P_1 \times \frac{T_2}{T_1} \).