One box containing 1 mole of He at 7-3 T0 and other box containing 1 mole of a polyatomic gas -1-33 at T0 are placed together to attain thermal equilibrium- The final temperature becomes Tf- Then -

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Standard XII

Physics

RMS and KE Expressions

One box conta...

Question

One box containing $1$ mole of $H e$ at $7 / 3 T_{0}$ and other box containing $1$ mole of a polyatomic gas $(γ = 1.33)$ at $T_{0}$ are placed together to attain thermal equilibrium. The final temperature becomes $T_{f}$ . Then :

T_{f} = \frac{9}{13} T_{0}

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T_{f} = \frac{13}{9} T_{0}

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T_{f} = \frac{5}{2} T_{0}

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T_{f} = \frac{3}{2} T_{0}

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Solution

The correct option is D $T_{f} = \frac{13}{9} T_{0}$
At thermal equilibrium, both have the same temperature and heat given by one gas equal to the heat absorbed by other gas.

Heat released by $H e =$ Heat taken up by polyatomic gas
$1 \times C_{v} \times Δ T = 1 \times C_{v} \times Δ T$
$1 \times \frac{3 R}{2} (7 / 3 T_{0} - T_{f}) = 1 \times 3 R \times (T_{f} - T_{0})$
$⎡ ⎢ ⎣ C_{v} = \frac{3}{2} R for He and C_{v} = \frac{3}{2} R + X = \frac{3}{2} R + \frac{3}{2} R = 3 R for polyatomic gas ⎤ ⎥ ⎦$
$∴ \frac{7}{2} T_{0} - \frac{3}{2} T_{f} = 3 T_{f} - 3 T_{0}$
$\frac{13}{2} T_{0} = \frac{9}{2} T_{f}$
$∴ T_{f} = \frac{13}{9} T_{0}$

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One box containing 1 mole of He at 7/3T0 and other box containing 1 mole of a polyatomic gas (γ=1.33) at T0 are placed together to attain thermal equilibrium. The final temperature becomes Tf. Then :

One box containing $1$ mole of $H e$ at $7 / 3 T_{0}$ and other box containing $1$ mole of a polyatomic gas $(γ = 1.33)$ at $T_{0}$ are placed together to attain thermal equilibrium. The final temperature becomes $T_{f}$ . Then :