The average kinetic energy of hydrogen molecules at \(300 \mathrm{~K}\) is \(E\). At the same temperature the average kinetic energy of oxygen molecules will be (A) [E/(16)] (B) \(E\) (C) \(4 \mathrm{E}\) D) \([E /(4)]\)

Short Answer

Expert verified
The average kinetic energy of oxygen molecules at \(300~K\) is the same as that of hydrogen molecules, which is represented by \(E\). So, the correct option is (B) \(E\).

Step by step solution

01

Identify the given information

We are given: 1. The temperature of both hydrogen and oxygen gases: \(T = 300~K\) 2. The average kinetic energy of hydrogen molecules: \(E\) Our goal is to find the average kinetic energy of oxygen molecules at the same temperature.
02

Recall the equation for the average kinetic energy

For an ideal gas, the average kinetic energy (\(E_k\)) of a molecule is given by: \[E_k = \dfrac{3}{2}kT\] where \(k\) is Boltzmann's constant, and \(T\) is the temperature in Kelvin.
03

Use the equation to determine the average kinetic energy of the hydrogen molecules

We are given that the average kinetic energy of hydrogen molecules at \(300~K\) is \(E\). So, using the equation, we can write: \[E = \dfrac{3}{2}kT\]
04

Determine the average kinetic energy of the oxygen molecules

Since the temperature is the same for both the hydrogen and oxygen gas molecules (\(T = 300~K\)), the average kinetic energies of oxygen molecules, \(E_o\), must also be given by: \[E_o = \dfrac{3}{2}kT\]
05

Compare the average kinetic energies of hydrogen and oxygen molecules

As per step 3 and 4, we know that \(E = \dfrac{3}{2}kT\) and \(E_o=\dfrac{3}{2}kT\). Comparing both expressions, we get: \[E_o = E\] So, the average kinetic energy of oxygen molecules at the same temperature is the same as that of hydrogen molecules.
06

Answer

The correct option is: (B) \(E\) The average kinetic energy of oxygen molecules at \(300~K\) is the same as that of hydrogen molecules, which is represented by \(E\).

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