An adiabatic vessel of total volume V is partitioned into two equal parts by a fixed, conducting separator. The left part holds one mole of an ideal gas ( U = 1.5 nRT ) and the right part contains two moles of the same

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Correct answer: A. (pV)/(3R)

Let the initial temperatures of the left and right parts be

T_L

and

T_R

respectively.

The volume of each part is

\dfrac{V}{2}

.

For the left part:

p \left(\dfrac{V}{2}\right) = (1) R T_L \Rightarrow T_L = \dfrac{pV}{2R}

For the right part:

p \left(\dfrac{V}{2}\right) = (2) R T_R \Rightarrow T_R = \dfrac{pV}{4R}

The internal energy of the gas is given by

U = 1.5 nRT = \dfrac{3}{2} nRT

.

The initial total internal energy of the system is:

U_i = \dfrac{3}{2} (1) R T_L + \dfrac{3}{2} (2) R T_R

U_i = \dfrac{3}{2} R

\left(\dfrac{pV}{2R}\right) +

3R \left(\dfrac{pV}{4R}\right)<br><br>

U_i =

\dfrac{3pV}{4} + \dfrac{3pV}{4} = \dfrac{3pV}{2}<br><br>Let the final common temperature be

T_f

. The final total internal energy is:<br><br>

U_f =

\dfrac{3}{2} (

n_1 + n_2) R T_f =

\dfrac{3}{2} (

1 + 2) R T_f =

\dfrac{9R T_f}{2}<br><br>Since the vessel is adiabatic and the separator is fixed, no heat is exchanged with the surroundings and no work is done. Therefore, the total internal energy of the system is conserved.<br><br>

U_i = U_f

<br><br>\dfrac{3pV}{2} = \dfrac{9R T_f}{2}<br><br>

T_f =

\dfrac{pV}{3R}

An adiabatic vessel of total volume V is partitioned into two equal...

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