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JEE Main 20206 SeptemberMorning Shift
Exam
JEE Main
Year
2020
Date
6 September
Shift
Morning Shift
Subject
Physics
Chapter
Thermodynamics

JEE Main 2020 · 6 September · Morning Shift · Physics · Thermodynamics

Initially a gas of diatomic molecules is contained in a cylinder of...

Initially a gas of diatomic molecules is contained in a cylinder of volume V1{V}_{1} at a pressure P1{P}_{1} and temperature 250K250K. Assuming that 2525 % of the molecules get dissociated causing a change in number of moles. The pressure of the resulting gas at temperature 2000K,2000K, when contained in a volume 2V12{V}_{1} is given by P2{P}_{2}. The ratio P2/P1{P}_{2} / {P}_{1} is -

Correct answer

5

Step-by-step solution

P1=P1{P}_{1} = {P}_{1}
v1=v1{v}_{1} = {v}_{1}
T1=250{T}_{1} = 250
n1=n0{n}_{1} = {n}_{0}
n2=0.75n0+2×0.25n0=1.25no{n}_{2} = 0 . 75{n}_{0} + 2 \times 0 . 25{n}_{0} = 1 . 25no
v2=2V1{v}_{2} = 2{V}_{1}
T2=2000{T}_{2} = 2000
P1V1=n1RT1{P}_{1}{V}_{1} = {n}_{1}{RT}_{1}
P2V2=n2RT2{P}_{2}{V}_{2} = {n}_{2}{RT}_{2}
P1V1=n0R×250{P}_{1}{V}_{1} = {n}_{0}R \times 250
P2(2V1)=(.25n0R(2000)){P}_{2}\left(2{V}_{1}\right) = \left(. 25{n}_{0}R ( 2000 )\right)
equation (1)/(2)( 1 ) / ( 2 )
P12P2=2501.25×2000\dfrac{{P}_{1}}{2{P}_{2}} = \dfrac{250}{1 . 25 \times 2000}
P1P2=15\dfrac{{P}_{1}}{{P}_{2}} = \dfrac{1}{5}
P2P1=5\dfrac{{P}_{2}}{{P}_{1}} = 5

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