JEE Main · 2006 · 30 Apr 2006 · Offline · Heat & Thermodynamics

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Two rigid boxes containing different ideal gases are placed on a table. Box A contains one mole of nitrogen at temperature

T_{0}

, while Box B contains one mole of helium at temperature (7/3)

T_{0}

. The boxes are then put into thermal contact with each other and heat flows between them until the gases reach a common final temperature. (Ignore the heat capacity of boxes). Then, the final temperature of the gases,

T_{f}

, in terms of

T_{0}

(A) $T_{f} = \frac{5}{2} T_{0}$
(B) $T_{f} = \frac{3}{7} T_{0}$
(C) $T_{f} = \frac{7}{3} T_{0}$
(D) $T_{f} = \frac{3}{2} T_{0}$ ✓ Correct

Answer: (D)

Solution

No work is done, so the heat given up by the hotter gas equals the heat taken up by the cooler one:

n_{A} C_{v, A} (T_{f} - T_{0}) + n_{B} C_{v, B} (T_{f} - \frac{7}{3} T_{0}) = 0

. Nitrogen is diatomic with

C_{v} = \frac{5}{2} R

and helium monatomic with

C_{v} = \frac{3}{2} R

, one mole of each. Thus

\frac{5}{2} (T_{f} - T_{0}) + \frac{3}{2} (T_{f} - \frac{7}{3} T_{0}) = 0

, which gives

8 T_{f} = 12 T_{0}

, so

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

JEE Main · 2006 · 30 Apr 2006 · Offline · Single correct option

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JEE Main · 2006 · 30 Apr 2006 · Offline · Heat & Thermodynamics

Solution

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