Problem 107 - Master Resource Book in JEE Main Physics

Question

The average translatory energy and rms speed of molecules in a sample of oxygen gas at \(300 \mathrm{~K}\) are \(5.21 \times 10^{-21} \mathrm{~J}\) and \(484 \mathrm{~ms}^{-1}\) respectively. The sorresponding values at \(600 \mathrm{~K}\) are nearly (assuming deal gas behaviour) (a) \(\left.12.42 \times 10^{-21}\right\rfloor, 968 \mathrm{~ms}^{-1}\) (b) \(7.78 \times 10^{-21} \mathrm{~J}, 684 \mathrm{~ms}^{-1}\) (c) \(6.21 \times 10^{-21} \mathrm{~J}, 968 \mathrm{~ms}^{-1}\) (d) \(12.42 \times 10^{-21} \mathrm{~J}, 684 \mathrm{~ms}^{-1}\)

Step-by-Step Solution

Answer

The calculated energy is incorrect; rms speed at 600 K is 684 ms^{-1}. Option d has correct rms speed.

Step 1: Understanding the formula for translational energy

The average translatory kinetic energy (E) of gas molecules at temperature (T) can be calculated using the formula: \[ E = \frac{3}{2}kT \]where \(k\) is the Boltzmann constant \(1.38 \times 10^{-23} \mathrm{~JK}^{-1}\). Since the energy is proportional to temperature in Kelvin, if the initial energy at \(300 \mathrm{~K}\) is \(5.21 \times 10^{-21} \mathrm{~J}\), doubling the temperature to \(600 \mathrm{~K}\) should double the energy.

Step 2: Calculating the translational energy at 600 K

Since the energy at \(300 \mathrm{~K}\) is \(5.21 \times 10^{-21} \mathrm{~J}\), the energy at \(600 \mathrm{~K}\) can be calculated by direct proportion:\[ E_{600} = 2 \times 5.21 \times 10^{-21} \mathrm{~J} = 10.42 \times 10^{-21} \mathrm{~J} \].

Step 3: Understanding the formula for rms speed

The root-mean-square (rms) speed (v) of the gas molecules is given by the formula: \[ v = \sqrt{\frac{3kT}{m}} \]where \(m\) is the molar mass of the gas, which remains constant. The rms speed is proportional to the square root of the temperature, as \(v \propto \sqrt{T}\).

Step 4: Calculating the rms speed at 600 K

The initial rms speed is \(484 \mathrm{~ms}^{-1}\) at \(300 \mathrm{~K}\). At \(600 \mathrm{~K}\), which is double the temperature, the new speed is:\[ v_{600} = 484 \times \sqrt{\frac{600}{300}} = 484 \times \sqrt{2} \approx 684 \mathrm{~ms}^{-1} \].

Step 5: Matching calculated values with options

From our calculations, at \(600 \mathrm{~K}\), the translational energy is \(10.42 \times 10^{-21} \mathrm{~J}\), and the rms speed is approximately \(684 \mathrm{~ms}^{-1}\). Among the given options, \( (d) 12.42 \times 10^{-21} \mathrm{~J}, 684 \mathrm{~ms}^{-1} \) has the correct rms speed, but incorrect energy. Hence, correct values do not directly match any options, suggesting possible typographical or miscalculation in options.

Key Concepts

Translational Kinetic Energy

In the realm of thermodynamics, translational kinetic energy is a fundamental concept that relates to the motion of particles. When we talk about gases, this energy is the motion of the molecules moving in space without spinning or vibrating. The formula to compute the average translational kinetic energy of a gas molecule is:\[ E = \frac{3}{2} kT \]In this equation, \(E\) represents the energy, \(k\) is the Boltzmann constant \(1.38 \times 10^{-23} \mathrm{~JK}^{-1}\), and \(T\) is the absolute temperature in Kelvins.

• Translational kinetic energy is directly proportional to temperature.
• As temperature doubles, the translational kinetic energy also doubles.

Breaking it down, the temperature effect means that if you increase the temperature of a gas, each molecule will have more energy. This is precisely what happens when the temperature rises from 300K to 600K, doubling the energy reflected in the formula calculation.

RMS Speed

Root-mean-square speed is an important parameter when discussing gas molecules' movement. The rms speed, often denoted as \(v\), provides insight into the distribution of molecular speeds at a given temperature. It can be calculated using the formula:\[ v = \sqrt{\frac{3kT}{m}} \]Here, \(k\) is again the Boltzmann constant, \(T\) is the temperature, and \(m\) is the mass of a single molecule. For practical calculations, the molar mass of the gas is often used.

• The rms speed is directly related to the square root of temperature.
• Raising the temperature results in an increase in rms speed according to the relation \(v \propto \sqrt{T}\).

Thus, when the temperature changes from 300K to 600K, the rms speed doesn't double like translational energy does; instead, it increases by a factor of \(\sqrt{2}\), demonstrating the non-linear relationship between temperature and rms speed.

Ideal Gas Law

The ideal gas law is a cornerstone of thermodynamics that relates the macroscopic properties of an ideal gas. It is expressed as:\[ PV = nRT \]In this equation:

• \(P\) is the pressure of the gas.
• \(V\) is the volume.
• \(n\) represents the number of moles.
• \(R\) is the ideal gas constant.
• \(T\) is the absolute temperature measured in Kelvins.

The ideal gas law provides the framework to understand how gas particles behave under various conditions. It's particularly useful for relating how changes in one property, like temperature, affect others such as pressure or volume, assuming a constant number of moles.While the ideal gas law doesn't directly calculate translational kinetic energy or rms speed, it provides a background context where these properties interact. For instance, increasing temperature (\(T\) in the ideal gas law) will subsequently affect energy and speed as calculated in other formulas.