Question

You have a sample of gas at \(-33^{\circ} \mathrm{C}\) . You wish to increase the rms speed by a factor of \(2 .\) To what temperature should the gas be heated?

Short answer

To double the rms speed of a gas initially at -33°C, you need to heat it to approximately 687.45°C.

Step-by-step solution

Convert initial temperature to Kelvin

First, we need to convert the initial temperature from Celsius to Kelvin. To do this, add 273.15 to the Celsius temperature: Initial temperature in Kelvin (\(T_1\)) = -33°C + 273.15 = 240.15 K

Set up the proportion

Now we will set up the proportion using the equation for rms speed. We have: \(v_{1_{rms}} = \sqrt{\frac{3R(T_1)}{M}}\) \(v_{2_{rms}} = 2v_{1_{rms}} = \sqrt{\frac{3R(T_2)}{M}}\) Since we want to find the final temperature (\(T_2\)), we can square both sides of the equation and eliminate the M and R variables as they are constants and will cancel out: \((2v_{1_{rms}})^2 = \frac{3R(T_2)}{M}\)

Replace \(v_{1_{rms}}\) with its expression

Replace \(v_{1_{rms}}\) in the equation with its expression in terms of the initial temperature (\(T_1\)): \((2\sqrt{\frac{3R(T_1)}{M}})^2 = \frac{3R(T_2)}{M}\)

Solve for \(T_2\)

Now, solve for the final temperature (\(T_2\)): \((2\sqrt{\frac{3R(T_1)}{M}})^2 = 4\frac{3R(T_1)}{M} = \frac{3R(T_2)}{M}\) Multiply both sides by M and divide by 3R: \(4(T_1) = T_2\) Now, substitute the value of \(T_1\) and solve for \(T_2\): \(T_2 = 4(240.15) = 960.6\, K\)

Convert final temperature to Celsius

Finally, convert the final temperature to Celsius: Final temperature in Celsius = 960.6 K - 273.15 = 687.45°C The gas should be heated to approximately 687.45°C to double its rms speed.

Key concepts

Gas Temperature Conversion

Understanding gas temperature conversion is essential when working with gas laws and their associated equations. The Kelvin scale is the standard unit of measure in these cases because it starts at absolute zero, the point where theoretically no more thermal energy can be extracted from a substance.

To convert from Celsius to Kelvin, which is a typical task in gas-related problems, you simply add 273.15 to the Celsius temperature. This adjustment allows for temperature to be accurately reflected in equations that are based on absolute temperature scales. It's crucial because the behavior of gases relates directly to their kinetic energy, which is a function of absolute temperature.

For instance, in the solution provided, the initial gas temperature of (-33 degrees Celsius) was converted to Kelvin ((240.15 K) by adding 273.15. When dealing with gas properties and dynamic reactions, converting to the Kelvin scale ensures that all temperature inputs reflect the absolute thermal energy present in the system, allowing for proper application of the kinetic molecular theory.

Root-Mean-Square Velocity

Root-mean-square (RMS) velocity is a way to express the average velocity of gas molecules in a sample. It gives us insight into the speed of gas particles and how they change with temperature. The RMS velocity is derived from the Kinetic Molecular Theory and is directly proportional to the square root of the absolute temperature. Mathematically, the RMS speed of a particle in a gas is given by the formula: \(v_{rms} = \sqrt{\frac{3kT}{m}}\).

Where \(v_{rms}\) is the root-mean-square velocity, 'T' is the absolute temperature in Kelvins, 'm' is the mass of a particle, and 'k' is the Boltzmann constant. This formula tells us that if you want to increase the RMS speed of gas particles, as in the exercise given, you need to increase the absolute temperature of the gas. Doubling the RMS speed requires quadrupling the absolute temperature, because of the square root relationship. This is why, when increasing the initial temperature of our gas sample by a factor of four, we achieve the desired increase in RMS speed.

Kinetic Molecular Theory of Gases

The Kinetic Molecular Theory of gases provides a conceptual framework for understanding the properties of gases and how they react to changes in the environment. This theory posits that gases are composed of a large number of particles, which are in constant, random motion. The assumptions of the theory include that these particles are point masses with no volume and do not interact with each other except during elastic collisions.

The temperature of a gas is a measure of the average kinetic energy of its particles. This means that as the temperature of a gas increases, the average speed of the particles also increases. This theory lays the foundation for why temperature plays a key role in gas dynamics. For example, as highlighted in the solution of the exercise, to double the RMS speed of the gas, which is linked to the kinetic energy of its particles, we must significantly increase the temperature. The exercise illustrates that understanding the kinetic molecular theory is integral in determining the behavior of gases under various conditions.