JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Heat & Thermodynamics

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A container is divided into two chambers by a partition . The volume of first chamber is 4.5 litre and second chamber is 5.5 litre. The first chamber contain 3.0 moles of gas at pressure 2.0 atm and second chamber contain 4.0 moles of gas at pressure 3.0 atm . After the partition is removed and the mixture attains equilibrium, then, the common equilibrium pressure existing in the mixture is

x \times 1 0^{- 1} atm

. Value of

x

is ____

Answer: 25.5

Solution

The two samples start at different temperatures,

T_{1} = \frac{P _{1} V _{1}}{n _{1} R}

and

T_{2} = \frac{P _{2} V _{2}}{n _{2} R}

. Removing the partition does no external work, and no heat leaves the container, so the total internal energy is unchanged. For an ideal gas

U = \frac{f}{2} n R T = \frac{f}{2} P V

, and with the same

f

for both samples this gives

P (V_{1} + V_{2}) = P_{1} V_{1} + P_{2} V_{2}

P = \frac{( 2.0 ) ( 4.5 ) + ( 3.0 ) ( 5.5 )}{10.0} = \frac{9.0 + 16.5}{10.0} = 2.55 atm

Since

2.55

atm is

25.5 \times 1 0^{- 1}

atm, the value of

x

25.5

JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Numerical answer

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JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Heat & Thermodynamics

Solution

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