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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

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) \(3 \mathrm{~V}\)

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) \(8: 9\)

To what temperature should the hydrogen at \(327^{\circ} \mathrm{C}\) cooled at constant pressure, so that the root mean square velocity of its molecules become half of its previous value (A) \(-100^{\circ} \mathrm{C}\) (B) \(123^{\circ} \mathrm{C}\) (C) \(0^{\circ} \mathrm{C}\) (D) \(-123^{\circ} \mathrm{C}\)

For a gas \(\gamma=(7 / 5)\), the gas may probably be (A) Neon (B) Argon (C) Helium (D) Hydrogen

Under constant temperature, graph between \(\mathrm{p}\) and \((1 / \mathrm{V})\) is (A) Hyperbola (B) Circle (C) Parabola (D) Straight line
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