JEE Main · 2020 · 06 Sep 2020 · Shift 1 · Heat & Thermodynamics

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

V_{1}

at a pressure

P_{1}

and temperature 250 K . Assuming that

25%

of the molecules get dissociated causing a change in number of moles. The pressure of the resulting gas at temperature 2000 K , when contained in a volume

2 V_{1}

is given by

P_{2}

. The ratio

\frac{P _{2}}{P _{1}}

is ____.

Answer: 5

Solution

Let

n_{0}

moles of diatomic gas be present initially, so

P_{1} V_{1} = n_{0} R (250)

. When a quarter of the molecules dissociate, each dissociating molecule gives two atoms: the surviving diatomic molecules number

0.75 n_{0}

and the atoms formed number

2 (0.25 n_{0}) = 0.5 n_{0}

, for a total of

n^{'} = 1.25 n_{0}

moles. Applying the gas law to the final state,

P_{2} (2 V_{1}) = 1.25 n_{0} R (2000)

. Dividing this by

P_{1} V_{1} = n_{0} R (250)

gives

\frac{2 P _{2}}{P _{1}} = \frac{1.25 \times 2000}{250} = 10

, so

\frac{P _{2}}{P _{1}} = 5

JEE Main · 2020 · 06 Sep 2020 · Shift 1 · Numerical answer

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JEE Main · 2020 · 06 Sep 2020 · Shift 1 · Heat & Thermodynamics

Solution

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