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Chapter 9: Problem 1262

The ratio of mean kinetic energy of hydrogen and nitrogen at temperature $300 \mathrm{~K}\( and \)450 \mathrm{~K}$ respectively is (A) \(2: 3\) (B) \(3: 2\) C) \(4: 9\) D) \(2: 2\)

Short Answer

Expert verified
The ratio of mean kinetic energies of hydrogen and nitrogen is calculated using the formula \(E_k = \dfrac{3}{2}RT\). Given the temperatures of \(300 \mathrm{~K}\) for hydrogen and \(450 \mathrm{~K}\) for nitrogen, we find the ratio to be \(\dfrac{E_{k,H}}{E_{k,N}} = \dfrac{300}{450} = \dfrac{2}{3}\). Thus, the correct answer is (A) \(2: 3\).

Step by step solution

01

Recall the formula for mean kinetic energy

The mean kinetic energy (\(E_k\)) of a gas molecule is given by the following formula: \[E_k = \dfrac{3}{2}RT\] Where R is the universal gas constant and T is the temperature in Kelvin.
02

Calculate the mean kinetic energy for both gases separately

Hydrogen gas and nitrogen gas have different molar masses, but since mean kinetic energy depends only on the temperature and the gas constant R (which is the same for both gases), we can simply calculate the mean kinetic energy for hydrogen and nitrogen just by plugging in the given temperature values. For hydrogen: \[E_{k,H} = \dfrac{3}{2}RT_H\] \[E_{k,H} = \dfrac{3}{2}R (300)\] For nitrogen: \[E_{k,N} = \dfrac{3}{2}RT_N\] \[E_{k,N} = \dfrac{3}{2}R (450)\]
03

Calculate the ratio of mean kinetic energies

Now we can directly calculate the ratio of mean kinetic energies of hydrogen to nitrogen: \[\dfrac{E_{k,H}}{E_{k,N}} = \dfrac{\dfrac{3}{2}R (300)}{\dfrac{3}{2}R (450)}\] We notice that the gas constant R and the constant factor 3/2 cancel out in the numerator and the denominator: \[\dfrac{E_{k,H}}{E_{k,N}} = \dfrac{300}{450}\] Now, we simplify the ratio: \[\dfrac{E_{k,H}}{E_{k,N}} = \dfrac{2}{3}\] So the ratio of the mean kinetic energy of hydrogen to nitrogen is 2:3. The correct answer is (A) \(2: 3\).

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Most popular questions from this chapter

The temperature of an ideal gas is increased from \(27^{\circ} \mathrm{C}\) to \(927^{\circ} \mathrm{C}\). The root mean square speed of its molecules becomes (A) Four times (B) One-fourth (C) Half (D) Twice

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