JEE Main 2017 · Physics · Thermodynamics and Kinetic Theory of Gases · hard

A thermally insulated rigid vessel is divided into two compartments by a closed valve. The first, of volume 2 L, holds a diatomic ideal gas at 3 atm…

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A thermally insulated rigid vessel is divided into two compartments by a closed valve. The first, of volume

2\ \mathrm{L}

, holds a diatomic ideal gas at

3\ \mathrm{atm}

and

300\ \mathrm{K}

. The second, of volume

3\ \mathrm{L}

, holds the same gas at

2\ \mathrm{atm}

and

400\ \mathrm{K}

. After the valve is opened and equilibrium is reached, the temperature of the gas is nearly

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Answer: (A) $343\ \mathrm{K}$

The vessel is rigid and insulated, so no work is done and no heat enters: the total internal energy is conserved. For the same gas,

U = nC_VT

, so

n_1T_1 + n_2T_2 = (n_1+n_2)T_f

. Using

nT = PV/R

n_1T_1 = 3\times2/R = 6/R

and

n_2T_2 = 2\times3/R = 6/R

. The mole numbers are

n_1 = 6/(300R)

and

n_2 = 6/(400R)

, so

T_f = \dfrac{12}{6/300 + 6/400} = \dfrac{12}{0.035} \approx 343\ \mathrm{K}

. The simple mean

350\ \mathrm{K}

ignores the unequal mole numbers.

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