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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

    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:

    A

    50%

    B

    40%

    C

    100%

    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 %

    Byju's Answer Standard XII Chemistry

    Gay Lussac's Law, Avagadro's Law

    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 %

    Gay Lussac's Law, Avagadro's Law

    Standard XII Chemistry

    Suggest Corrections 4 Video Solution

    JEE - Grade 11 - Chemistry - States of Matter - Session 03 - W11

    CHEMISTRY 01:49 Min | 5 Views Rate

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    स्रोत : byjus.com

    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

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    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.

    \Rightarrow\qquad \frac{P_2}{P_1}=\frac{V_1}{4V_1}\times \frac{2T_1}{T_

    \Rightarrow\qquad \frac{P_2}{P_1}=\frac{V_1}{4V_1}\times \frac{2T_1}{T_

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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

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    Mohammed 2 month ago
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