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31. (AIEEE 2008 )

An insulated container of gas has two chambers separated by an insulating partition. One of the chambers has volume V 1 and contains ideal gas at pressure P 1 and temperature T 1 . The other chamber has volume V 2 and contains ideal gas at pressure P 2 and temperature T 2 . If the partition is removed without doing any work on the gas, the final equilibrium temperature of the gas in the container will be

(A) T 1 T 2 ( P 1 V 1 + P 2 V 2 ) P 1 V 1 T 2 + P 2 V 2 T 1

(B) P 1 V 1 T 1 + P 2 V 2 T 2 P 1 V 1 + P 2 V 2

(C) P 1 V 1 T 2 + P 2 V 2 T 1 P 1 V 1 + P 2 V 2

(D) T 1 T 2 ( P 1 V 1 + P 2 V 2 ) P 1 V 1 T 1 + P 2 V 2 T 2

Correct answer is (A)

Since, no work is done and system is thermally insulated from surrounding. Therefore, total internal energy is constant, that is, U = U 1 + U 2 Assuming gas have same degree of freedom, we have

( n 1 + n 2 ) C V T = n 1 C V T 1 + n 2 C V T 2

Therefore,

T = n 1 R T 1 + n 2 R T 2 R n 1 + n 2 R = ( P 1 V 1 + P 2 V 2 ) P 1 V 1 T 1 + P 2 V 2 T 2 = ( P 1 V 1 + P 2 V 2 ) T 1 T 2 ( P 1 V 1 T 2 + P 2 V 2 T 1 )

32. (AIEEE 2004 )

Two thermally insulated vessels 1 and 2 are filled with air at temperatures ( T 1 , T 2 ) , volume ( V 1 , V 2 ) and pressure ( P 1 , P 2 ) respectively. If the value joining the two vessels is opened, the temperature inside the vessel at equilibrium will be

(A) T 1 T 2 ( P 1 V 1 + P 2 V 2 ) ( P 1 V 1 T 2 + P 2 V 2 T 1 )

(B) ( T 1 + T 2 ) / 2

(C) T 1 + T 2

(D) T 1 T 2 ( P 1 V 1 + P 2 V 2 ) ( P 1 V 1 T 1 + P 2 V 2 T 2 )

Correct answer is (A)

Here Q = 0 and W = 0. Therefore from first law of thermodynamics Δ U = Q + W = 0
Internal energy of the system with partition = Internal energy of the system without partition.
n 1 C v T 1 + n 2 C v T 2 = ( n 1 + n 2 ) C v T
T = n 1 T 1 + n 2 T 2 n 1 + n 2
But n 1 = P 1 V 1 R T 1 and n 2 = P 2 V 2 R T 2
T = P 1 V 1 R T 1 × T 1 + P 2 V 2 R T 2 × T 2 P 1 V 1 R T 1 + P 2 V 2 R T 2
= T 1 T 2 ( P 1 V 1 + P 2 V 2 ) P 1 V 1 T 2 + P 2 V 2 T 1