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# the temperature of a certain mass of a gas is quadrupled. if the gas initially is at 2 atm pressure, find the % increase in pressure.

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## Initially, a gas is present at 2 atm pressure. If the temperature of the gas is doubled keeping amount of gas and volume constant, then the percentage increase in pressure will be:

Click here👆to get an answer to your question ✍️ Initially, a gas is present at 2 atm pressure. If the temperature of the gas is doubled keeping amount of gas and volume constant, then the percentage increase in pressure will be: Question

A

B

C

D

## 80%

Medium Open in App

Updated on : 2022-09-05

Solution Verified by Toppr

Correct option is C)

P∝T⇒

T 1 ​ P 1 ​ ​ = T 2 ​ P 2 ​ ​ ⇒ T 2 ​ = 2T P ​ ⇒P 2 ​ =4atm

Percentage increase=

2 (4−2) ​ ×100=100%

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## The temperature of a certain mass of a gas is doubled. If the initial pressure of the gas is 1 atm, find the % increase in pressure.A. 50 %B. 100 %C. 150 %D. 200 %

The temperature of a certain mass of a gas is doubled. If the initial pressure of the gas is 1 atm, find the % increase in pressure.A. 50 %B. 100 %C. 150 %D. 200 % The temperatu... Question

The temperature of a certain mass of a gas is doubled. If the initial pressure of the gas is 1 atm, find the % increase in pressure.

A 50% B 100% C 150% D 200% Open in App Solution

The correct option is B 100%

Given information:

Initial pressure of the gas

P 1 = 1 a t m

Initial temperature of the gas

T 1 = T K

Final temperature of the gas

T 2 = 2 T K

Final Pressure of the gas

P 2 = ?

From Gay Lussac's law,

P 1 T 1 = P 2 T 2 at constant volume. 1 T = P 2 2 T Hence, P 2 = 2 a t m

Increase in pressure

= 2 − 1 = 1 atm

Percentage increase in pressure

=

increase in pressure

initial pressure × 100 = 1 1 × 100 = 100 %

Standard XII Chemistry

Suggest Corrections 4 Video Solution JEE - Grade 11 - Chemistry - States of Matter - Session 03 - W11

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## What is the pressure of a gas if the temperature of the gas is doubled and the volume is quadrupled?

Answer (1 of 8): The correct answer to this question depends on the order in which you conduct these changes. Thus, without more information I cannot help. I suggest you go back to your teacher and ask him/her for the additional information needed to complete this assignment. What is the pressure of a gas if the temperature of the gas is doubled and the volume is quadrupled?

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Juan Eduardo Ynsil Alfaro

Studies at National University of Engineering (Universidad Nacional de Ingeniería-UNI) (Expected 2024)Author has 251 answers and 179.9K answer views3y

The pressure is inversely proportional to the volume (Boyle’s Law) and directly proportional to the temperature (Gay Lussac’s Law)

So it will double and divide by four. The result is that it halves.

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Sarthak Gupta

B. Tech in Computer Science and Engineering, Indian Institute of Information Technology, Pune (IIITP) (Expected 2023)3y

Pressure becomes half (provided the gas is ideal).

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Kailash Mehra

Chemical Engineer - Plant Design & Six Sigma Black BeltAuthor has 455 answers and 169.5K answer views1y

For a ideal gas pressure, temperature and volume is related by below relationship.

P = R*T/V

Here, R is gas constant.

So if we do Consider revised state pressure by P1 then,

P1 = R*2T/4V P1/P = 1/2 P1 = P/2

So pressure will be 1/2 of the original state..

Gopal Menon

B Tech in Chemical Engineering, Indian Institute of Technology, Bombay (IITB) (Graduated 1975)Upvoted by

Bob Spillman

, PhD Chemistry, University of Illinois at Urbana-Champaign (1975)Author has 10.1K answers and 9.8M answer views3y

What is the pressure of a gas if the temperature of the gas is doubled and the volume is quadrupled?

For an ideal gas, PV=nRT, PV=nRT, where P,V,n,R P,V,n,R and T T

are the pressure, volume, number of moles of gas present, gas constant and temperature respectively.

Hence, if the number of moles of the gas is constant,

PV T PVT is constant. ⇒ P 1 V 1 T 1 = P 2 V 2 T 2 ⇒P1V1T1=P2V2T2 for constant n. n. ⇒ P 2 P 1 = V 1 T 2 V 2 T 1 = V 1 V 2 × T 2 T 1 .

⇒P2P1=V1T2V2T1=V1V2×T2T1.

It is given that T 2 =2 T 1 T2=2T1 and V 2 =4 V 1 . V2=4V1.

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Chris VdBerg

Management Consultant at Project & Business Consulting (2015–present)1y

The relationship is as following:

P1 * V1 / T1 =P2 * V2 / T2

P2 = P1 * V1 / V2 / T1 * T2

P2 = P1 * V1 / (4 V1) / T1 * (T1 * 2)

P2 = P1 (2 / 4) P2 = 1/2 * P1

The new pressure (P2) is half the initial pressure (P1)

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Dennis Leppin

B.E.and M.E. in Chemical Engineering, City College of New YorkAuthor has 1.7K answers and 899.7K answer views3y

These changes are linear for ideal gases. Non ideality aside, this should halve the pressure.

Bill Wilson (she/it)

Consultant at Self-Employment (2017–present)Author has 2.3K answers and 306.6K answer viewsJul 1

The correct answer to this question depends on the order in which you conduct these changes. Thus, without more information I cannot help. I suggest you go back to your teacher and ask him/her for the additional information needed to complete this assignment.

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David Wilmshurst

Materials Engineering Mgr - Concrete & Structural TestingAuthor has 8.2K answers and 11M answer views3y

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Mohammed 2 month ago

Guys, does anyone know the answer?