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Two containers (A and B) are in thermal contact with an environment at temperature T = 280 K. The two containers are connected by a tube that contains an ideal turbine (a mechanical device) and a valve. Initially, the valve is closed. Container A (volume VA = 2 m3) is filled with n = 1.2 moles of an ideal monatomic gas, and container B (volume VB = 3.5 m3) is empty (vacuum). When the valve is opened, gas flows between A and B, and the turbine is used to generate electricity. What is the maximum amount of work than can be done on the turbine as the gas reaches equilibrium? Remember: U = 1.5 nRT + const S = nR ln(V) + f(U,n)

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okpalawalter8

Answer:

The maximum amount of work is  [tex]W = 1563.289 \ J[/tex]

Explanation:

From  the question we are told that

   The temperature of the environment is  [tex]T = 280\ K[/tex]

    The volume of container A is  [tex]V_A = 2 m^3[/tex]

    Initially the number of moles  is  [tex]n = 1.2 \ moles[/tex]

     The volume of container B is [tex]V_B = 3.5 \ m^3[/tex]

     

At equilibrium of the gas the maximum work that can be done on the turbine is mathematically represented as

             [tex]W = P_A V_A ln[ \frac{V_B}{V_A} ][/tex]

Now from the Ideal gas law

          [tex]P_A V_A = nRT[/tex]

So substituting for [tex]P_A V_A[/tex] in the equation above

          [tex]W = nRT ln [\frac{V_B}{V_A} ][/tex]

Where R is the gas constant with a values of  [tex]R = 8.314 \ J/mol[/tex]

Substituting values we have that

            [tex]W = 1.2 * (8.314) * (280) * ln [\frac{3.5}{2} ][/tex]

          [tex]W = 1563.289 \ J[/tex]