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Chapter 9: Problem 1256

The relation between the gas pressure \(\mathrm{P}\) and average kinetic energy per unit volume \(E\) is (A) \(\mathrm{P}=(2 / 3) \mathrm{E}\) (B) \(P=(3 / 2) E\) (C) \(\mathrm{P}=\mathrm{E}\) (D) \(\mathrm{P}=(\mathrm{E} / 2)\)

Short Answer

Expert verified
The correct relation between gas pressure P and average kinetic energy per unit volume E is given by the formula \(P = \frac{2}{3}E\). This is derived from the ideal gas law and the expression for average kinetic energy per unit volume. The correct answer is (A).

Step by step solution

01

Recall the ideal gas law

The ideal gas law is given by the equation: \[P V = n R T\] where P is the gas pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvins.
02

Define average kinetic energy per unit volume

The average kinetic energy per unit volume is given by the equation: \[E = \frac{(\frac{3}{2})nRT}{V}\] where E is the average kinetic energy per unit volume.
03

Find the relation between P and E

Now we want to find the relation between the gas pressure P and the average kinetic energy per unit volume E. To do this, we'll use the expressions we have derived in Step 1 and Step 2. From the ideal gas law (Step 1), \[P = \frac{nRT}{V}\] Also, we have the expression for E from Step 2, \[E = \frac{(\frac{3}{2})nRT}{V}\] Divide the expression for E by the expression for P, \[\frac{E}{P} = \frac{(\frac{3}{2})nRT}{V} \times \frac{V}{nRT}\] This simplifies to \[\frac{E}{P} = \frac{3}{2}\] Now, we can rearrange the equation to find the relation between P and E: \[P = \frac{2}{3}E\] So the correct answer is (A) \(\mathrm{P}=(2 / 3) \mathrm{E}\).

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Most popular questions from this chapter

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) \(1: 1\)

If the ratio of vapor density for hydrogen and oxygen is \([1 /(16)]\), then under constant pressure the ratio of their \(\mathrm{rms}\) velocities will be (A) \(4: 1\) (B) \(1: 16\) (C) \(16: 1\) (D) \(1: 4\)

At \(\mathrm{O}^{\circ} \mathrm{C}\) the density of a fixed mass of a gas divided by pressure is \(\mathrm{x} .\) At \(100^{\circ} \mathrm{C}\), the ratio will be (A) \(\mathrm{x}\) (B) \([(273) /(373)] \mathrm{x}\) (C) \([(373) /(273)] \mathrm{x}\) (D) \([(100) /(273)] \mathrm{x}\)

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) \(\mathrm{C}_{\mathrm{V}}-\mathrm{C}_{\mathrm{P}}=\mathrm{J}\)

If the molecular weight of two gases are \(\mathrm{M}_{1}\) and \(\mathrm{M}_{2}\), then at a given temperature the ratio of root mean square velocity \(\mathrm{v}_{1}\) and \(v_{2}\) will be (A) \(\sqrt{\left(\mathrm{M}_{1} / \mathrm{M}_{2}\right)}\) (B) \(\sqrt{\left(M_{2} / M_{1}\right)}\) (C) $\left.\sqrt{[}\left(\mathrm{M}_{1}-\mathrm{M}_{2}\right) /\left(\mathrm{M}_{1}+\mathrm{M}_{2}\right)\right]$ (D) $\left.\sqrt{[}\left(\mathrm{M}_{1}+\mathrm{M}_{2}\right) /\left(\mathrm{M}_{1}-\mathrm{M}_{2}\right)\right]$
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