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Chapter 9: Problem 1269

The average translational kinetic energy of \(\mathrm{O}_{2}\) (molar mass 32 ) molecules at a particular temperature \(0.068 \mathrm{eV}\). The translational kinetic energy of \(\mathrm{N}_{2}\) (molar mass 28 ) molecules in \(\mathrm{eV}\) at the same temperature is (A) \(0.003 \mathrm{eV}\) (B) \(0.068 \mathrm{eV}\) (C) \(0.056 \mathrm{eV}\) (D) \(0.678 \mathrm{eV}\)

Short Answer

Expert verified
The average translational kinetic energy of N₂ molecules at the same temperature as O₂ molecules is (C) $0.056 \mathrm{eV}$.

Step by step solution

01

Recall the formula for the average translational kinetic energy

We know that the average translational kinetic energy of a molecule in a gas (K.E) is given by the formula: \[K.E = \frac{3}{2} \cdot kT\] where \(k\) is the Boltzmann constant, and \(T\) is the temperature in Kelvin.
02

Obtain the relationship between K.E and molar mass

We can write the given formula in terms of molar mass \(M\) as follows: \[K.E = \frac{3}{2} \cdot R \cdot T \cdot \frac{1}{M}\] where \(R\) is the universal gas constant.
03

Calculate the temperature

We have been given the average translational kinetic energy of O₂ (K.E O₂) as \(0.068\: eV\). Convert it to Joules: \[K.E (\text{O₂}) = 0.068 \: eV \times 1.6 \times 10^{-19} J/eV\] We also know the molar mass of O₂ is 32 g/mol. Substitute these values in the formula in step 2 and solve for the temperature \(T\): \[0.068 \times 1.6 \times{10}^{-19} = \frac{3}{2} \cdot R \cdot T \cdot \frac{1}{32}\]
04

Calculate the average translational kinetic energy of N₂

Now that we have the temperature, we can calculate the average translational kinetic energy of N₂ using the molar mass of N₂, which is 28 g/mol. Substitute the temperature in the formula in step 2: \[K.E(\text{N₂}) = \frac{3}{2} \cdot R \cdot T \cdot \frac{1}{28}\]
05

Convert the kinetic energy from Joules to eV

Finally, we need to convert the average translational kinetic energy of N₂ from Joules to electron volts (eV) using the conversion factor: \[\text{Kinetic energy (eV)} = \text{Kinetic energy (J)} \div 1.6 \times 10^{-19}\]
06

Identify the correct option

Compare the calculated value of the average translational kinetic energy of N₂ in eV with the given options (A, B, C, and D) and select the one that matches our calculated value. By following these steps and calculations, we will find the average translational kinetic energy of N₂ molecules at the same temperature as O₂ molecules.

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Most popular questions from this chapter

\(2 \mathrm{~g}\) of \(\mathrm{O}_{2}\) gas is taken at \(27^{\circ} \mathrm{C}\) and pressure \(76 \mathrm{~mm} \mathrm{Hg}\). Find out volume of gas (ln liter) (A) \(3.08\) (B) \(44.2\) (C) \(2.05\) (D) \(2.44\)

At what temperature pressure remaining constant will the \(\mathrm{rms}\) speed of a gas molecules increase by \(10 \%\) of the \(\mathrm{rms}\) speed at NTP? (A) \(57.3 \mathrm{~K}\) (B) \(57.3^{\circ} \mathrm{C}\) (C) \(557.3 \mathrm{~K}\) (D) \(-57.3^{\circ} \mathrm{C}\)

A sample of gas is at \(0^{\circ} \mathrm{C}\). To what temperature it must be raised in order to double the rms speed of molecule. (A) \(270^{\circ} \mathrm{C}\) (B) \(819^{\circ} \mathrm{C}\) (C) \(100^{\circ} \mathrm{C}\) (D) \(1090^{\circ} \mathrm{C}\)

The temperature of an ideal gas is increased from \(27^{\circ} \mathrm{C}\) to \(127^{\circ} \mathrm{C}\), then percentage increase in \(\mathrm{v}_{\mathrm{rms}}\) is (A) \(33 \%\) (B) \(11 \%\) (C) \(15.5 \%\) (D) \(37 \%\)

A perfect gas at \(27^{\circ} \mathrm{C}\) is heated at constant pressure to \(327^{\circ} \mathrm{C}\). If original volume of gas at \(27^{\circ} \mathrm{C}\) is \(\mathrm{V}\) then volume at \(327^{\circ} \mathrm{C}\) is (A) \(2 \mathrm{~V}\) (B) \(\mathrm{V}\) (C) (V/2) (D) \(3 \mathrm{~V}\)
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