Thermodynamics

JEE Main level, numerical answer, medium

132 of 190

Question 132 of 190

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 -

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Answer: 5

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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