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Chapter 9: Problem 1301

Calculate the temperature at which \(\mathrm{rms}\) velocity of \(\mathrm{SO}_{2}\) molecules is the same as that of \(\mathrm{O}_{2}\) molecules at \(27^{\circ} \mathrm{C}\). Molecular weights of Oxygen and \(\mathrm{SO}_{2}\) are \(32 \mathrm{~g}\) and \(64 \mathrm{~g}\) respectively (A) \(327^{\circ} \mathrm{C}\) (B) \(327 \mathrm{~K}\) (C) \(127^{\circ} \mathrm{C}\) (D) \(227^{\circ} \mathrm{C}\)

Short Answer

Expert verified
The temperature at which the rms velocity of SO2 molecules is the same as that of O2 molecules at 27°C is \(327^{\circ}\mathrm{C}\).

Step by step solution

01

Write down the expression for rms velocity for both gases

The expression for rms velocity (v_rms) is given by the formula: \(v_{rms} = \sqrt{\dfrac{3RT}{M}}\) Where: - R is the universal gas constant, - T is the temperature in Kelvin - M is the molar mass of the gas. For O2 molecules at 27°C or 300K (273 + 27), we can write the expression for v_rms as: \(v_{rms_{O2}} = \sqrt{\dfrac{3R \cdot 300}{32}}\) For SO2 molecules, we want to find the temperature where their rms velocity is equal to the rms velocity of O2 molecules. So we can write: \(v_{rms_{SO2}} = \sqrt{\dfrac{3R \cdot T_{SO2}}{64}}\) Now, we need to find the temperature \(T_{SO2}\) where the rms velocities for both gases are equal.
02

Set the rms velocities equal to each other and solve for the temperature

Since we want to find the temperature where the rms velocities for both gases are equal, we can set their expressions equal to each other: \(\sqrt{\dfrac{3R \cdot 300}{32}} = \sqrt{\dfrac{3R \cdot T_{SO2}}{64}}\) Now, we need to solve for \(T_{SO2}\) by squaring both sides and then isolating the variable on one side: \(\dfrac{3R \cdot 300}{32} = \dfrac{3R \cdot T_{SO2}}{64}\) Let's isolate \(T_{SO2}\) by multiplying both sides by 64, dividing by 3R, and simplifying: \(T_{SO2} = \dfrac{300 \cdot 64}{32} = 600\) So the temperature we found is in Kelvin. To convert it into Celsius, we need to subtract 273: \(T_{SO2(\mathrm{Celsius})} = 600 - 273 = 327^{\circ}\mathrm{C}\) The correct answer is option (A).

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Most popular questions from this chapter

Suppose ideal gas equation follows \(V P^{3}=\) constant, Initial temperature and volume of the gas are \(\mathrm{T}\) and \(\mathrm{V}\) respectively. If gas expand to \(27 \mathrm{~V}\), then temperature will become (A) \(9 \mathrm{~T}\) (B) \(27 \mathrm{~T}\) (C) (T/9) (D) \(\mathrm{T}\)

The volume of a gas at \(20 \mathrm{C}\) is \(200 \mathrm{ml}\). If the temperature is reduced to \(-20^{\circ} \mathrm{C}\) at constant pressure, its volume will be. (A) \(172.6 \mathrm{~m} 1\) (B) \(17.26 \mathrm{ml}\) (C) \(19.27 \mathrm{ml}\) (D) \(192.7 \mathrm{ml}\)

The average translational energy and \(\mathrm{rms}\) speed of molecules in sample of oxygen gas at \(300 \mathrm{~K}\) are $6.21 \times 10^{-21} \mathrm{~J}\( and \)484 \mathrm{~m} / \mathrm{s}$ respectively. The corresponding values at \(600 \mathrm{~K}\) are nearly (assuming ideal gas behavior) (A) \(6.21 \times 10^{-21} \mathrm{~J}, 968 \mathrm{~m} / \mathrm{s}\) (B) \(12.42 \times 10^{-21} \mathrm{~J}, 684 \mathrm{~m} / \mathrm{s}\) (C) \(12.42 \times 10^{-21} \mathrm{~J}, 968 \mathrm{~m} / \mathrm{s}\) (D) \(8.78 \times 10^{-21} \mathrm{~J}, 684 \mathrm{~m} / \mathrm{s}\)

When the pressure on \(1200 \mathrm{ml}\) of a gas is increased from $70 \mathrm{~cm}\( to \)120 \mathrm{~cm}$ of mercury at constant temperature, the new volume of the gas will be (A) \(400 \mathrm{ml}\) (B) \(600 \mathrm{ml}\) (C) \(700 \mathrm{ml}\) (D) \(500 \mathrm{ml}\)

At what temperature, pressure remaining unchanged, will the rms velocity of a gas be half its value at \(0^{\circ} \mathrm{C}\) ? (A) \(204.75 \mathrm{~K}\) (B) \(204.75^{\circ} \mathrm{C}\) (C) \(-204.75 \mathrm{~K}\) (D) \(-204.75^{\circ} \mathrm{C}\)
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