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Two rigid boxes {1}{2} containing different ideal gases are pl

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Two rigid boxes $\frac{1}{2}$ 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 $T_{0}$ is
(a) $T_{f} = \frac{3}{7} T_{0}$
(b) $T_{f} = \frac{7}{3} T_{0}$
(c) $T_{f} = \frac{3}{2} T_{0}$
(d) $T_{f} = \frac{5}{2} T_{0}$

Name: Video solution 1: Two rigid boxes 21 containing different ideal gases are placed on a table. Box A contains one mole of nitrogen at temperature T0, while box B contains one mole of helium at temperature (7/3) T0. 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, Tf, in terms of T0 is(a) Tf=73T0(b) Tf=37T0(c) Tf=23T0(d) Tf=25T0
Uploaded: 2023-12-25T18:28:05.000Z
Duration: 5 min

Views: 5,252 students

Updated on: Dec 25, 2023

Book:

Arihant Physics JEE Main Chapterwise Solutions (2019-2002) (Arihant)

Exam:

JEE Mains 2006

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Mateo Discussed

Two rigid boxes

\frac{1}{2}

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}

is
(a)

T_{f} = \frac{3}{7} T_{0}

(b)

T_{f} = \frac{7}{3} T_{0}

(c)

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

(d)

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

15 mins ago

Discuss this question LIVE

15 mins ago

Text solutionVerified

E x p . (c)

Here, change in internal energy of the system is zero, i.e., increase in internal energy of one is equal to decrease in internal energy of other.
Solution Image

Change in internal energy in box

A

Δ U_{A} = 1 \times \frac{5}{2} R (T_{f} - T_{0})

Change in internal energy in box

B

Δ U_{B} = 1 \times \frac{3 R}{2} (T_{f} - \frac{7}{3} T_{0})

Now,

Δ U_{A} + Δ U_{B} = 0

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

5 T_{f} - 5 T_{0} + 3 T_{f} - 7 T_{0} = 0

8 T_{f} = 12 T_{0}

T_{f} = \frac{12}{8} T_{0} = \frac{3}{2} T_{0}

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Question Text	Two rigid boxes $\frac{1}{2}$ 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 $T_{0}$ is (a) $T_{f} = \frac{3}{7} T_{0}$ (b) $T_{f} = \frac{7}{3} T_{0}$ (c) $T_{f} = \frac{3}{2} T_{0}$ (d) $T_{f} = \frac{5}{2} T_{0}$
Updated On	Dec 25, 2023
Topic	Thermodynamics
Subject	Physics
Class	Class 11
Answer Type	Text solution:1 Video solution: 1
Upvotes	176
Avg. Video Duration	5 min

Text solutionVerified

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