Question

A balloon contains \(2500 \mathrm{~mL}\) of helium gas at \(75^{\circ} \mathrm{C}\). What is the new volume, in milliliters, of the gas when the temperature changes to each of the following, if \(n\) and \(P\) do not change? a. \(55^{\circ} \mathrm{C}\) b. \(680 . \mathrm{K}\) c. \(-25^{\circ} \mathrm{C}\) d. \(240 . \mathrm{K}\)

Short answer

a) 2357.89 mL; b) 4882.48 mL; c) 1781.58 mL; d) 1723.55 mL.

Step-by-step solution

- Use Combined Gas Law

Start with the combined gas law formula for situations where the amount of gas and pressure are constant: \[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \] Here, \(V_1\) is the initial volume (2500 mL), \(T_1\) is the initial temperature in Kelvin, \(V_2\) is the final volume, and \(T_2\) is the final temperature in Kelvin.

- Convert Initial Temperature to Kelvin

Convert the initial temperature from Celsius to Kelvin using the formula: \[ T_{\text{K}} = T_{\text{C}} + 273.15 \] For the initial temperature of \(75^{\circ} \text{C}\): \[ T_1 = 75 + 273.15 = 348.15 \text{ K} \]

- Convert Given Temperatures to Kelvin

Convert each given temperature to Kelvin: a) For \(55^{\circ} \text{C}\): \[ T_2 = 55 + 273.15 = 328.15 \text{ K} \] b) \(680 \text{ K}\) is already in Kelvin. c) For \(-25^{\circ} \text{C}\): \[ T_2 = -25 + 273.15 = 248.15 \text{ K} \] d) \(240 \text{ K}\) is already in Kelvin.

- Calculate New Volumes

Use the combined gas law to solve for \(V_2\) in each case: a) For \(55^{\circ} \text{C}\): \[ \frac{2500}{348.15} = \frac{V_2}{328.15} \implies V_2 = \frac{2500 \times 328.15}{348.15} \approx 2357.89 \text{ mL} \] b) For \(680 \text{ K}\): \[ \frac{2500}{348.15} = \frac{V_2}{680} \implies V_2 = \frac{2500 \times 680}{348.15} \approx 4882.48 \text{ mL} \] c) For \(-25^{\circ} \text{C}\): \[ \frac{2500}{348.15} = \frac{V_2}{248.15} \implies V_2 = \frac{2500 \times 248.15}{348.15} \approx 1781.58 \text{ mL} \] d) For \(240 \text{ K}\): \[ \frac{2500}{348.15} = \frac{V_2}{240} \implies V_2 = \frac{2500 \times 240}{348.15} \approx 1723.55 \text{ mL} \]

Key concepts

Gas Volume Calculation

Understanding gas volume calculation is crucial in chemistry, especially when dealing with changing conditions like temperature or pressure. Here's a breakdown of how to calculate the new volume of a gas when its temperature changes, while keeping the amount of gas (n) and pressure (P) constant. The Combined Gas Law equation: \[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \] is what you use. Here, \( V_1 \) is the initial volume of the gas, and \( T_1 \) is the initial temperature in Kelvin. \( V_2 \) is the final volume that we need to find, and \( T_2 \) is the final temperature in Kelvin. Start by converting all temperatures to Kelvin. Using the Combined Gas Law, rearrange the equation to solve for \( V_2 \): \[ V_2 = V_1 \times \frac{T_2}{T_1} \] By substituting the known values into this equation, you will find the new volumes.

Temperature Conversion

Temperature conversion, particularly between Celsius and Kelvin, is essential when working with gas laws. Since gas law calculations require temperature in Kelvin, knowing how to convert from Celsius is a must. The formula to convert from Celsius to Kelvin is: \[ T_{\text{K}} = T_{\text{C}} + 273.15 \] For example, if you have an initial temperature of \(75^{\text{C}}\), convert it to Kelvin by adding 273.15, giving you 348.15 K. Repeat this step for every temperature provided in the problem. Remember, Kelvin is always used in gas law equations, so accuracy in these conversions is very important. This practice also helps in standardizing results and makes the calculations simpler.

Ideal Gas Law

The Ideal Gas Law is another fundamental concept when discussing gases. Though our primary focus here is on the Combined Gas Law, understanding the Ideal Gas Law provides a broader context. The equation for the Ideal Gas Law is: \[ PV = nRT \] Here, \( P \) stands for pressure, \( V \) is volume, \( n \) is the number of moles of gas, \( R \) is the gas constant, and \( T \) is temperature in Kelvin. Even though we're using the Combined Gas Law for this exercise, all these gas law equations interrelate. While solving the exercise, maintaining constant \( n \) and \( P \), allows the equation to simplify, highlighting the direct proportionality between volume and temperature. This reinforces why we use Kelvin for temperature and how these concepts are linked in gas calculations.