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Chapter 9: Problem 1302

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 \(200 \mathrm{~K}\)

Short Answer

Expert verified
The root-mean-square speed at 800 K is twice the value at 200 K. The correct choice is (B) Twice the value at 200 K.

Step by step solution

01

Write down the rms speed formula

The formula for the root-mean-square (rms) speed of a gas is given by: \(v_\text{rms} = \sqrt{\frac{3RT}{M}}\) where \(v_\text{rms}\) is the root-mean-square speed, \(R\) is the universal gas constant, \(T\) is the temperature of the gas, and \(M\) is the molar mass of the gas.
02

Determine the relationship between the rms speeds

Divide the rms speed formula at 800 K by the rms speed formula at 200 K to find the relationship between the two rms speeds: \(\frac{v_\text{rms, 800 K}}{v_\text{rms, 200 K}} = \frac{\sqrt{\frac{3R(800)}{M}}}{\sqrt{\frac{3R(200)}{M}}}\)
03

Simplify the equation

Notice that we can cancel out the factors of \(3R\) and \(M\) in the numerator and denominator: \(\frac{v_\text{rms, 800 K}}{v_\text{rms, 200 K}} = \frac{\sqrt{800}}{\sqrt{200}}\) Next, simplify the expression by taking the square root of 800 and 200: \(\frac{v_\text{rms, 800 K}}{v_\text{rms, 200 K}} = \frac{20\sqrt{2}}{10\sqrt{2}}\) Now, we can cancel out the factor of \(\sqrt{2}\) in the numerator and denominator: \(\frac{v_\text{rms, 800 K}}{v_\text{rms, 200 K}} = \frac{20}{10}\)
04

Compute the final answer

Divide 20 by 10 to find the final answer: \(\frac{v_\text{rms, 800 K}}{v_\text{rms, 200 K}} = 2\) The root-mean-square speed at 800 K is twice the value at 200 K. The correct choice is (B) Twice the value at 200 K.

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Most popular questions from this chapter

The relation between two specific heats of a gas is (A) $\mathrm{C}_{\mathrm{V}}-\mathrm{C}_{\mathrm{P}}=(\mathrm{R} / \mathrm{J})$ (B) \(\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}=\mathrm{J}\) (C) $\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}=(\mathrm{R} / \mathrm{J})$ (D) \(\mathrm{C}_{\mathrm{V}}-\mathrm{C}_{\mathrm{P}}=\mathrm{J}\)

The root mean square speed of hydrogen molecules of an ideal hydrogen kept in a gas chamber at \(0^{\circ} \mathrm{C}\) is \(3180 \mathrm{~ms}^{-1}\). The pressure on the hydrogen gas is (Density of hydrogen gas is $8.99 \times 10^{-2} \mathrm{~kg} / \mathrm{m}^{3}, 1 \mathrm{~atm}=1.01 \times 10^{5} \mathrm{Nm}^{-2}$ ) (A) \(1.0 \mathrm{~atm}\) (B) \(3.0 \mathrm{~atm}\) (C) \(2.0 \mathrm{~atm}\) (D) \(1.5 \mathrm{~atm}\)

A vessel contains 1 mole of \(\mathrm{O}_{2}\) gas (molar mass 32 ) at a temperature \(\mathrm{T}\). The pressure of the gas is \(\mathrm{P}\). An identical vessel containing one mole of He gas (molar mass 4) at temperature \(2 \mathrm{~T}\) has a pressure of (A) \(2 \mathrm{P}\) (B) \(\mathrm{P}\) (C) \(8 \mathrm{P}\) (D) \((\mathrm{P} / 8)\)

The pressure is exerted by the gas on the walls of the container because (A) It sticks with the walls (B) It is accelerated towards the walls (C) It loses kinetic energy (D) On collision with the walls there is a change in momentum

At a given temperature the root mean square velocities of Oxygen and hydrogen molecules are in the ratio (A) \(1: 4\) (B) \(1: 16\) (C) \(16: 1\) (D) \(4: 1\)
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