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Chapter 9: Problem 1261

The ratio of mean kinetic energy of hydrogen and oxygen at a given temperature is (A) \(1: 8\) (B) \(1: 4\) (C) \(1: 16\) (D) \(1: 1\)

Short Answer

Expert verified
The ratio of mean kinetic energy of hydrogen and oxygen at a given temperature is 1:16, represented by the expression \(\frac{M_{O_2}}{M_{H_2}}\), where \(M_{O_2}\) and \(M_{H_2}\) are the molar masses of oxygen and hydrogen, respectively. The correct answer is (C) \(1: 16\).

Step by step solution

01

Understand the relationship between mean kinetic energy, molecular mass, and temperature

According to the kinetic molecular theory, the mean kinetic energy of a gas is given by the expression: Mean Kinetic Energy (\(KE_{avg}\)) = \(\frac{3}{2} \times \frac{R \times T}{M}\) where R is the gas constant, T is the temperature in Kelvin and M is the molar weight of the gas (measured in kg/mol). We want to find the ratio of the mean kinetic energies of hydrogen and oxygen at a given temperature. The ratio can be written as: Ratio = \(\frac{KE_{H_2}}{KE_{O_2}}\)
02

Calculate the molar masses of hydrogen and oxygen

We need to determine the molar masses of hydrogen (H₂) and oxygen (O₂). The atomic weight of hydrogen (H) is approximately 1 g/mol, and since there are two hydrogen atoms in one molecule of hydrogen (H₂), the molar mass of hydrogen (M_H₂) is approximately 2 g/mol or 0.002 kg/mol. For oxygen, the atomic weight of oxygen (O) is approximately 16 g/mol, and since there are two oxygen atoms in one molecule of oxygen (O₂), the molar mass of oxygen (M_O₂) is approximately 32 g/mol or 0.032 kg/mol.
03

Find the ratio of mean kinetic energies

Now that we have the molar masses of hydrogen and oxygen, we can find the ratio of mean kinetic energies by substituting the molar masses into the expression for the ratio: Ratio = \(\frac{KE_{H_2}}{KE_{O_2}} = \frac{\frac{3}{2} \times \frac{R \times T}{M_{H_2}}}{\frac{3}{2} \times \frac{R \times T}{M_{O_2}}}\) Notice that both the temperature (T) and the gas constant (R) are given, so we can cancel them out: Ratio = \(\frac{M_{O_2}}{M_{H_2}}\) Now, we can substitute the values of the molar masses: Ratio = \(\frac{0.032}{0.002}\) Ratio = 16 So, the ratio of mean kinetic energies of hydrogen to oxygen at the given temperature is 1:16.
04

Choose the correct answer

From our calculation, we found that the ratio of mean kinetic energies of hydrogen to oxygen at a given temperature is 1:16. Therefore, the correct answer is: (C) \(1: 16\)

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Most popular questions from this chapter

For a gas, the rms speed at \(800 \mathrm{~K}\) is (A) Four times the value at \(200 \mathrm{~K}\) (B) Twice the value at \(200 \mathrm{~K}\) (C) Half the value at \(200 \mathrm{~K}\) (D) same as at \(200 \mathrm{~K}\)

At constant temperature on increasing the pressure of a gas \(5 \%\) its volume will decrease by (A) \(5 \%\) (B) \(5.26 \%\) (C) \(4.76 \%\) (D) \(4.26 \%\)

Root mean square velocity of a molecule is \(v\) at pressure \(P\). If pressure is increased two times, then the rms velocity becomes (A) \(3 \mathrm{~V}\) (B) \(2 \mathrm{v}\) (C) \(0.5 \mathrm{~V}\) (D) \(\mathrm{v}\)

At what temperature is the kinetic energy of a gas molecule double that of its value at \(27^{\circ} \mathrm{C}\) (A) \(54^{\circ} \mathrm{C}\) (B) \(108^{\circ} \mathrm{C}\) (C) \(327^{\circ} \mathrm{C}\) (D) \(300^{\circ} \mathrm{C}\)

The ratio of the vapor densities of two gases at a given temperature is $9: 8$, The ratio of the rms velocities of their molecule is (A) \(3: 2 \sqrt{2}\) (B) \(2 \sqrt{2}: 3\) (C) \(9: 8\) (D) \(8: 9\)
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