Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 9: Problem 1227

\- decrease the volume of gas by \(\supset \%\) at constant temperature the pressure should be (A) Increased by \(5.26 \%\) (B) Decreased by \(5.26 \%\) (C) Decreased by \(11 \%\) (D) Increased by \(11 \%\)

Short Answer

Expert verified
The pressure of the gas will increase by approximately \(11 \%\) when the volume is decreased by \(10 \%\) at constant temperature. The correct answer is (D) Increased by \(11 \%\).

Step by step solution

01

Recall Boyle's Law

Boyle's Law states that Pressure(P) is inversely proportional to Volume(V) at constant temperature(T). Mathematically, it can be represented as: \[P \propto \frac{1}{V}\]
02

Express the relationship as an equation

To solve the problem, eliminate the proportionality and add a constant (k). So, we have: \[ P = \frac{k}{V}\] where k is the proportionality constant.
03

Determine the initial and final volumes

Since the volume is decreased by 10%, the initial volume (V1) can be represented as: \[V1 = V\] And the final volume (V2) can be represented as: \[V2 = 0.9V\] where V represents the original volume of the gas.
04

Apply Boyle's Law for initial and final states

By applying Boyle's Law to both the initial and final states, we have following equations: For initial state, \[P1 = \frac{k}{V}\] For final state, \[P2 = \frac{k}{0.9V}\]
05

Eliminate the constant k

We can eliminate the constant k by dividing the equation of the final state by the equation of the initial state: \[\frac{P2}{P1} = \frac{\frac{k}{0.9V}}{\frac{k}{V}}\]
06

Solve for the ratio of final to initial pressure

Simplify the equation to find the ratio of the final pressure (P2) to the initial pressure (P1): \[\frac{P2}{P1} = \frac{V}{0.9V}\] Simplifying further, we get: \[\frac{P2}{P1} = \frac{1}{0.9}\]
07

Find the percentage increase in pressure

Now find the percentage increase in pressure by dividing the final pressure by the initial pressure and then subtracting 1: Percentage increase = \(\frac{P2 - P1}{P1} \times 100\%) Use the ratio P2/P1 to get: Percentage increase = \(\frac{\frac{1}{0.9} - 1}{1} \times 100\%\) Simplify further to obtain the percentage increase in pressure: Percentage increase = \(\frac{0.1}{0.9} \times 100\% \approx 11.11\%\) So, the pressure has increased by approximately 11.11%. The correct answer is (D) Increased by \(11 \%\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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) $\left.\sqrt{[}\left(\mathrm{M}_{1}+\mathrm{M}_{2}\right) /\left(\mathrm{M}_{1}-\mathrm{M}_{2}\right)\right]$

A diatomic gas molecule has translational, rotational and vibrational degrees of freedom. The $\left(\mathrm{C}_{\mathrm{p}} / \mathrm{C}_{\mathrm{V}}\right)$ is (A) \(1.29\) (B) \(1.33\) C) \(1.4\) (D) \(1.67\)

Hydrogen gas is filled in a balloon at \(20^{\circ} \mathrm{C}\). If temperature is made \(40^{\circ} \mathrm{C}\), pressure remaining the same what fraction of hydrogen will come out (A) \(0.75\) (B) \(0.07\) (C) \(0.25\) D) \(0.5\)

The volume of a gas at pressure \(21 \times 10^{4} \mathrm{Nm}^{-2}\) and temperature \(27^{\circ} \mathrm{C}\) is 83 Liters. If $\mathrm{R}=8.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$. Then the quantity of gas in \(\mathrm{g}\) -mole will be (A) 42 (B) 7 (C) 14 (D) 15

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) \(1.5 \mathrm{~atm}\)
See all solutions

Recommended explanations on English Textbooks

Discourse

Read Explanation

Rhetoric

Read Explanation

Essay Prompts

Read Explanation

Email

Read Explanation

Essay Writing Skills

Read Explanation

Creative Story

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free