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Chapter 9: Problem 1333

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

Short Answer

Expert verified
The relationship between pressure (\(p\)) and the inverse of volume (\(\frac{1}{V}\)) under constant temperature can be represented by the equation \(p = \frac{k}{V}\), which shows a direct relationship. Thus, the graph between \(p\) and \(\frac{1}{V}\) is a straight line. The correct answer is (D) Straight line.

Step by step solution

01

Boyle's law can be represented by the following equation: \[pV = k\] Where: - \(p\) is the pressure of the gas - \(V\) is the volume of the gas - \(k\) is a constant that depends on the amount of gas and the temperature As the temperature is kept constant, the value of \(k\) is also constant in this case. #Step 2: Rearrange the equation to show the relation between p and 1/V#

Since we need to find the graph between p and (1/V), we'll need to rewrite the equation in terms of these two variables. To do so, we take the inverse of both sides of the equation: \[\frac{1}{p}\frac{1}{V} = \frac{1}{k}\] Multiplying both sides of the equation by \(V\), we get: \[\frac{1}{p} = \frac{V}{k}\] Now, multiplying both sides of the equation by \(p\), we get the final equation: \[p = \frac{k}{V}\] #Step 3: Identify the type of graph between p and 1/V#
02

The equation that we obtained in the previous step is: \[p = \frac{k}{V}\] From this equation, we can see that the relationship between pressure and the inverse of volume is a direct relationship, as one goes up, the other also goes up. This is a characteristic of a straight line. Hence, the graph between pressure (\(p\)) and the inverse of volume (\(1/V\)) under constant temperature is a straight line. #Step 4: Choose the correct answer#

Based on our analysis, the correct answer is: (D) Straight line

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Most popular questions from this chapter

At \(100 \mathrm{~K}\) and \(0.1\) atmospheric pressure, the volume helium gas is 10 liters. If volume and pressure are doubled, its temperature will change to (A) \(127 \mathrm{~K}\) (B) \(400 \mathrm{~K}\) (C) \(25 \mathrm{~K}\) (D) \(200 \mathrm{~K}\)

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

The rms speed of gas molecules is given by (A) \(2.5 \sqrt{\left[M_{0} /(R T)\right]}\) (B) \(\left.2.5 \sqrt{[}(\mathrm{RT}) / \mathrm{M}_{0}\right]\) (C) \(\left.1.73 \sqrt{[}(\mathrm{RT}) / \mathrm{M}_{0}\right]\) (D) \(1.73 \sqrt{\left[M_{0} /(R T)\right]}\)

The speeds of 5 molecules of a gas (in arbitrary units) are as follows: \(2,3,4,5,6\), The root mean square speed for these molecules is (A) \(4.24\) (B) \(2.91\) (C) \(4.0\) (D) \(3.52\)

A gas at \(27^{\circ} \mathrm{C}\) temperature and 30 atmospheric pressure $1 \mathrm{~s}$ allowed to expand to the atmospheric pressure if the volume becomes two times its initial volume, then the final temperature becomes (B) \(-173^{\circ} \mathrm{C}\) (A) \(273^{\circ} \mathrm{C}\) (C) \(173^{\circ} \mathrm{C}\) (D) \(100^{\circ} \mathrm{C}\)
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