Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 8: Problem 1167

In an isochoric Process \(\mathrm{T}_{1}=27^{\circ} \mathrm{C}\) and \(\mathrm{T}_{2}=127^{\circ} \mathrm{C}\) then $\left(\mathrm{P}_{1} / \mathrm{P}_{2}\right)$ will be equal to (A) \((9 / 59)\) (B) \((2 / 3)\) (C) \((4 / 3)\) (D) \((3 / 4)\)

Short Answer

Expert verified
The ratio of the initial and final pressures (\(P_1 / P_2\)) in the given isochoric process is equal to \(\frac{3}{4}\).

Step by step solution

01

Convert temperatures to Kelvin

We are given the temperatures in degrees Celsius. To work with the ideal gas law, we need to convert these temperatures to Kelvin. To do this, we add 273.15 to each temperature: \(T_1 = 27^\circ C + 273.15 = 300.15 K\) \(T_2 = 127^\circ C + 273.15 = 400.15 K\)
02

Apply ideal gas law in isochoric process

In an isochoric process, the volume remains constant. The ideal gas law is given by: \(PV = nRT\) Since the volume is constant, we can write the ideal gas law for the initial and final states as: \(P_1 V = n R T_1\) \(P_2 V = n R T_2\) Here, V (volume), n (number of moles), and R (ideal gas constant) are constants.
03

Find the ratio of initial to final pressure

We want to find the ratio \(P_1 / P_2\). Divide the first equation by the second equation: \(\frac{P_1 V}{P_2 V} = \frac{n R T_1}{n R T_2}\) Since V, n, and R are constants, they cancel out: \(\frac{P_1}{P_2} = \frac{T_1}{T_2}\)
04

Plug in the temperatures in Kelvin

Now, we plug in the temperatures in Kelvin that we found in Step 1: \(\frac{P_1}{P_2} = \frac{300.15 K}{400.15 K}\)
05

Simplify the ratio

Simplify the ratio \(P_1 / P_2\): \(\frac{P_1}{P_2} = \frac{300.15}{400.15} = \frac{3}{4}\) The ratio of the initial and final pressures (\(P_1 / P_2\)) is equal to \(\frac{3}{4}\). Therefore, the correct answer is (D) \(\frac{3}{4}\).

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

Two cylinders \(\mathrm{A}\) and \(\mathrm{B}\) fitted with piston contain equal amounts of an ideal diatomic gas at \(300 \mathrm{k}\). The piston of \(\mathrm{A}\) is free to move, While that of \(\mathrm{B}\) is held fixed. The same amount of heat is given to the gas in each cylinders. If the rise in temperature of the gas in \(\mathrm{A}\) is \(30 \mathrm{~K}\), then the rise in temperature of the gas in \(\mathrm{B}\) is. (A) \(30 \mathrm{~K}\) (B) \(42 \mathrm{~K}\) (C) \(18 \mathrm{~K}\) (D) \(50 \mathrm{~K}\)

Work done by \(0.1\) mole of a gas at \(27^{\circ} \mathrm{C}\) to double its volume at constant Pressure is \(\quad\) Cal. $R=2\left\\{(\mathrm{Cal}) /\left(\mathrm{mol}^{\circ} \mathrm{K}\right)\right\\}$ (A) 600 (B) 546 (C) 60 (D) 54

What is the value of absolute temperature on the Celsius Scale? (A) \(-273.15^{\circ} \mathrm{C}\) (B) \(100^{\circ} \mathrm{C}\) (C) \(-32^{\circ} \mathrm{C}\) (D) \(0^{\circ} \mathrm{C}\)

Each molecule of a gas has \(\mathrm{f}\) degrees of freedom. The radio \(\left(\mathrm{C}_{\mathrm{p}} / \mathrm{C}_{\mathrm{v}}\right)=\gamma\) for the gas is (A) \(1+(\mathrm{f} / 2)\) (B) \(1+(1 / \mathrm{f})\) (C) \(1+(2 / \mathrm{f})\) (D) \(1+\\{(\mathrm{f}-1) / 3\\}\)

Instructions:Read the assertion and reason carefully to mask the correct option out of the options given below. (A) If both assertion and reason are true and the reason is the correct explanation of the assertion. (B) If both assertion and reason are true but reason is not be correct explanation of assertion. (C) If assertion is true but reason is false. (D) If the assertion and reason both are false. Assertion: The total translation kinetic energy of all the molecules of a given mass of an ideal gas is \(1.5\) times the product of its Pressure and its volume. Reason: The molecules of a gas collide with each other and velocities of the molecules change due to the collision (A) D (B) \(\mathrm{C}\) (C) A (D) B
See all solutions

Recommended explanations on English Textbooks

Morphology

Read Explanation

Listening and Speaking

Read Explanation

Pragmatics

Read Explanation

Language Acquisition

Read Explanation

Linguistic Terms

Read Explanation

Blog

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