JEE Main · 2020 · 06 Sep 2020 · Shift 1 · States of Matter

A spherical balloon of radius 3 cm containing helium gas has a pressure of

48 \times 1 0^{- 3}

bar. At the same temperature, the pressure, of a spherical balloon of radius 12 cm containing the same amount of gas will be

____

\times 1 0^{- 6}

bar.

Answer: 750

Solution

At fixed temperature and amount of gas,

P V

is constant, and for a sphere

V \propto r^{3}

, so

P r^{3}

stays the same between the two states:

P_{1} r_{1}^{3} = P_{2} r_{2}^{3}

. Thus

48 \times 1 0^{- 3} \times 3^{3} = P_{2} \times 1 2^{3}

, i.e.

48 \times 1 0^{- 3} \times 27 = P_{2} \times 1728

. Solving,

P_{2} = \frac{48 \times 1 0 ^{- 3}}{64} = 0.75 \times 1 0^{- 3} bar = 750 \times 1 0^{- 6} bar

, so the value is

750

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

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